$\displaystyle x^2+y^2+z^2 = k(xy+yz+zx)$
First, conceptualize the solution set as a set points $\displaystyle (x, y, z) \in \mathbb{R}^3$.
A direct check shows that the origin is always in the solution set for any value of $\displaystyle k$, so will hunt for non-origin solutions only.
The origin is obviously the only point in the solution for $\displaystyle k = 0$. So assume going forward that $\displaystyle k \ne 0$.
Next, the equation is homogeneous... if $\displaystyle (x, y, z)$ is a solution, then so is $\displaystyle (t x, t y, t z)$ for any $\displaystyle t \in \mathbb{R}$.
By homogeneity, given a non-origin point P that's a solution, every point on the line through the origin and P is a solution.
Thus the total solution set is entirely determined by the solutions which are on the unit sphere (and if none, then the solution set is just the origin).
The solution set the is union of the all points on the set of lines going through the origin and any point of the solution on the unit sphere.
Holding off on the unit sphere for a moment, use that $\displaystyle (x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)$ to eliminate the cross terms on the right-hand side.
Do that by adding $\displaystyle \left(\frac{k}{2}\right)(x^2 + y^2 + z^2)$ to both sides, getting
$\displaystyle \left(1 + \frac{k}{2}\right)(x^2 + y^2 + z^2) = \left(\frac{k}{2}\right)(x+y+z)^2$. Dividing both sides by $\displaystyle \frac{k}{2}$ (recall assuming $\displaystyle k>0$) gives:
$\displaystyle (x+y+z)^2 = \left( \frac{1 + \frac{k}{2}}{\frac{k}{2}}\right)(x^2 + y^2 + z^2)$, and so
(*)$\displaystyle (x+y+z)^2 = \left(\frac{2 + k}{k}\right)(x^2 + y^2 + z^2)$
Now by (*), the left hand side must be non-negative, as must right-hand side's factor $\displaystyle (x^2 + y^2 + z^2)$. Thus the only solution will be the origin when $\displaystyle \frac{2 + k}{k} < 0$.
The solution set of $\displaystyle \frac{2 + k}{k} < 0 \ (k \ne 0)$ is $\displaystyle -2 < k < 0$.
Thus the origin is the only solution for $\displaystyle -2 < k \le 0$.
Now, assume $\displaystyle k$ is such that $\displaystyle \frac{2 + k}{k} > 0$ (meaning $\displaystyle k \notin (-2, 0\]$ ), and consider just the points on the unit sphere ($\displaystyle x^2 + y^2 + z^2 = 1$).
Take the square root of (*) and it reads
$\displaystyle | \ x+y+z \ | = \sqrt{\frac{2 + k}{k}}$.
Letting $\displaystyle c_k = \sqrt{\frac{2 + k}{k}}$, that says that any solution point $\displaystyle (x_0, y_0, z_0)$ on the unit sphere must also be on one of the two planes $\displaystyle x+y+z= c_k$ and $\displaystyle x+y+z = -c_k$.
Now if a solution $\displaystyle (x_0, y_0, z_0)$ on the unit sphere (so $\displaystyle x_0^2 + y_0^2 + z_0^2 = 1$), then by homogenity, its antipodal point $\displaystyle (-x_0, -y_0, -z_0)$ is also a solution on the unit sphere.
So if that point is also on the plane $\displaystyle x+y+z= c_k$ (meaning $\displaystyle x_0+y_0+z_0= c_k$), then $\displaystyle (-x_0, -y_0, -z_0)$ is a solution, and that satisfies $\displaystyle -x_0 - y_0 - z_0 = -c_k$.
Likewise, if a solution $\displaystyle (x_0, y_0, z_0)$ is on the unit sphere and in the plane $\displaystyle x+y+z= -c_k$, then its antipodal point $\displaystyle (-x_0, -y_0, -z_0)$ is a solution on the unit sphere and in the plane $\displaystyle x+y+z= c_k$.
Therefore, to find all non-origin solutions, we only need to find the intersection of unit sphere and the plane $\displaystyle x+y+z= c_k$, since at the end we'll build all solutions by including all the points on all the lines through the origin and a point on that intersection. The solutions on the unit sphere intersect the plane $\displaystyle x+y+z = -c_k$ will be captured in that process.
Note that the case $\displaystyle k = -2$ gives $\displaystyle c_k = 0$ is the case where those planes, $\displaystyle x+y+z = -c_k$ and $\displaystyle x+y+z = -c_k$, are the same plane... they've come together and met. Everything applies in that case as well.
In $\displaystyle \mathbb{R}^3$, a plane intersect the unit sphere is either the empty set (no intersection), a single point (the plane is tangent to the sphere), or a circle (the plane slices through the sphere).
Thus if you draw the all lines through the origin the go through a plane intersect the unit sphere, you'll either get the empty set, or a single line, or a double cone.
Thus the final solution to the problem will either be (depending on $\displaystyle k$): just the origin (since the origin is always a solution!), or a single line through the origin, or a double cone through the origin.
Now work out the precise solutions for a given $\displaystyle k$:
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Again, consider $\displaystyle k \notin \(-2, 0\]$, and $\displaystyle c_k = \sqrt{\frac{2 + k}{k}}$, and look only for the points on the unit sphere intersect the plane $\displaystyle x+y+z= c_k$.
One of that plane's normal vectors is $\displaystyle \vec{n} = (1, 1, 1)$. Let $\displaystyle \hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1)$. Notice that this doesn't vary with $\displaystyle k$.
Let the plane's point closest point to the origin be defined as $\displaystyle \vec{q}_k$.
Then $\displaystyle \vec{q}_k = R_k \hat{n}$ for some $\displaystyle R_k$. (That's geometrically obvious and has a straightforward proof.)
Whether, and/or how, the plane intersects the unit sphere is determined entirely by $\displaystyle |R_k|$.
If $\displaystyle |R_k| > 1$, then there's no intersection. If $\displaystyle |R_k| = 1$, then there's a single point of intersection (the plane's tangent to the sphere). If $\displaystyle |R_k|<1$, then the intersection is a circle.
To find $\displaystyle R_k$, use that $\displaystyle \vec{q}_k = R_k \hat{n} = R_k \left(\frac{1}{\sqrt{3}} (1, 1, 1)\right) = \frac{R_k}{\sqrt{3}} (1, 1, 1)$ is a point on the plane $\displaystyle x+y+z= c_k$.
Thus $\displaystyle \left( \frac{R_k}{\sqrt{3}} \right) + \left( \frac{R_k}{\sqrt{3}} \right) + \left( \frac{R_k}{\sqrt{3}} \right) = c_k$, so $\displaystyle 3\left( \frac{R_k}{\sqrt{3}} \right)= c_k$, so
$\displaystyle \sqrt{3} R_k = c_k$, $\displaystyle \sqrt{3} R_k = \sqrt{\frac{2 + k}{k}}$, so $\displaystyle R_k = \sqrt{\frac{2 + k}{3k}}$. (Note that $\displaystyle R_k \ge 0$, as the geometric picture suggested.)
With the formula $\displaystyle R_k = \sqrt{\frac{2 + k}{3k}} (> 0)$ (remember it's always assumed that $\displaystyle k \notin \(-2, 0\]$), can now classify solutions according to $\displaystyle k$.
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There's no intersection when $\displaystyle R_k > 1$, hence whenever $\displaystyle \sqrt{\frac{2 + k}{3k}} > 1$. Solving that for k gives $\displaystyle 0 < k < 1$, and in that case the grand solution to the problem is just the origin.
That can be rolled into the "origin only" solution when $\displaystyle k \notin (-2, 0\]$, so that: for $\displaystyle -2 < k < 1$, the only solution to the main problem is just the origin ($\displaystyle x = y = z = 0$).
Thus the grand solution is the origin only in the case where $\displaystyle -2 < k < 1$
$\displaystyle x = 0, y = 0, z = 0$
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There's only one point of intersection when $\displaystyle R_k = 1$, so when $\displaystyle \sqrt{\frac{2 + k}{3k}} = 1$. That occurs only when $\displaystyle k = 1$.
The single point of intersection between the unti sphere and the plane in that case is the point $\displaystyle \vec{q}_1 = R_1 \hat{n} = (1) \hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1) $.
Therefore the solution set when $\displaystyle k = 1$ is the line through the origin and that point $\displaystyle \vec{q}_1 = \frac{1}{\sqrt{3}} (1, 1, 1)$, so is the line $\displaystyle \left\{ \ \vec{q}_1 \tilde{t} \in \mathbb{R}^3 \ | \tilde{t} \in \mathbb{R} \right\}$.
Letting $\displaystyle t = \frac{\tilde{t}}{\sqrt{3}}$, that can be written parametrically without using the square root.
Thus the grand solution, a line, in the case where $\displaystyle k=1$, is given by
$\displaystyle x(t) = t, y(t) = t, z(t) = t, t \in \mathbb{R}$, or even more concisely written as the line x = y = z
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The intersection is a circle when $\displaystyle 0 \le R_k<1$, so when $\displaystyle \sqrt{\frac{2 + k}{3k}} < 1$ and $\displaystyle k \notin \(-2, 0\]$, so when either $\displaystyle k>1$ or $\displaystyle k \le -2$.
In the case $\displaystyle R_k<1$, the circle in 3-space formed by the plane intersect the unit sphere, will have center $\displaystyle \vec{q}_k$, and a normal vector $\displaystyle \hat{n}$ (since it's on the plane with that normal vector).
Let $\displaystyle r_k$ be the radius of that circle. Then by looking at the triangle from the origin, to $\displaystyle \vec{q}_k = R_k \hat{n}$, to a point on the circle, it's a right triangle with hypotenuses 1 and legs $\displaystyle |R_k|, r_k$.
Therefore $\displaystyle r_k = \sqrt{1 - R_k^2} = \sqrt{1 - \left(\sqrt{\frac{2 + k}{3k}}\right)^2} = \sqrt{1 - \frac{2 + k}{3k}} = \sqrt{\frac{2k - 2}{3k}}$$\displaystyle = \sqrt{\frac{2}{3}}\sqrt{\frac{k - 1}{k}}$.
To find unit vectors mutually orthogonal to $\displaystyle \hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1)$, find one unit vector orthogonal to it, and then take the cross product to find another (they'll form an orthonormal basis for $\displaystyle \mathbb{R}^3$).
It's easy to find one vector orthogonal to it, $\displaystyle (1, -1, 0)$, so let $\displaystyle \hat{u} = \frac{1}{\sqrt{2}} (1, -1, 0)$.
Then let $\displaystyle \hat{v} = \hat{n} \times \hat{u} = \frac{1}{\sqrt{6}}(1, 1, -2)$.
The points on the circle can be parameterized using sines and cosines, except instead of the classic x and y axes of $\displaystyle \mathbb{R}^2$, it's using the $\displaystyle \hat{u}$ and $\displaystyle \hat{v}$ "axes" in $\displaystyle \mathbb{R}^3$.
To get to a point on that circle, go to $\displaystyle \vec{q}_k = R_k \hat{n}$ and then move radius-times-cosine in the $\displaystyle \hat{u}$ (like classic x-axis) direction, then radius-times-sine in the $\displaystyle \hat{y}$ (like classic y-axis) direction.
The points on that circle will be given by: $\displaystyle R_k \hat{n} + r_k \cos(\theta) \hat{u} + r_k \sin(\theta) \hat{v}$ for all $\displaystyle \theta \in \mathbb{R}$.
In coordinates, that's $\displaystyle \frac{R_k}{\sqrt{3}} (1, 1, 1) + \frac{r_k \cos(\theta)}{\sqrt{2}} (1, -1, 0) + \frac{r_k \sin(\theta)}{\sqrt{6}}(1, 1, -2)$, so
$\displaystyle x = \frac{R_k}{\sqrt{3}} + \frac{r_k \cos(\theta)}{\sqrt{2}} + \frac{r_k \sin(\theta)}{\sqrt{6}}$
$\displaystyle = \sqrt{\frac{2 + k}{k}} + \left(\frac{\cos(\theta)}{\sqrt{2}} + \frac{\sin(\theta)}{\sqrt{6}}\right)\sqrt{\frac{2} {3}}\sqrt{\frac{k - 1}{k}}$
$\displaystyle = \sqrt{\frac{2 + k}{k}} + \left(\frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3}\right) \sqrt{\frac{k - 1}{k}}$.
$\displaystyle y = \frac{R_k}{\sqrt{3}} - \frac{r_k \cos(\theta)}{\sqrt{2}} + \frac{r_k \sin(\theta)}{\sqrt{6}}$
$\displaystyle = \sqrt{\frac{2 + k}{k}} + \left( - \frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3} \right) \sqrt{\frac{k - 1}{k}}$.
$\displaystyle z = \frac{R_k}{\sqrt{3}} - 2 \frac{r_k \sin(\theta)}{\sqrt{6}}$
$\displaystyle = \sqrt{\frac{2 + k}{k}} - \frac{2 \sin(\theta)}{3}\sqrt{\frac{k - 1}{k}}$.
Thus the grand solution, the double cone, in the case where $\displaystyle k>1$ or $\displaystyle k \le -2$, is given by
$\displaystyle x(t, \theta, k) = \left\[ \sqrt{\frac{2 + k}{k}} + \left(\frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3}\right) \sqrt{\frac{k - 1}{k}}\right\] t$
$\displaystyle y(t, \theta, k) = \left\[ \sqrt{\frac{2 + k}{k}} + \left( - \frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3} \right) \sqrt{\frac{k - 1}{k}}\right\] t$
$\displaystyle z(t, \theta, k) = \left\[ \sqrt{\frac{2 + k}{k}} - \frac{2 \sin(\theta)}{3}\sqrt{\frac{k - 1}{k}}\right\] t$