For what value of k will the following set of planes intersect in a line?
x-2y-z=0
x+9y-5z=0
kx-y+z=0
Hint: To find the line of intersection of the planes $\displaystyle x-2y-z=0$ and $\displaystyle x+9y-5z=0$, use the Addition Method to find the projected line on the $\displaystyle xy$-plane:
$\displaystyle \begin{aligned}
5x-10y-5z &= 0 \;\;\;\;\;\;\;\;\;\;\mbox{(multiply first equation by 5)}\\
x+9y-5z &= 0 \\
4x-19y &= 0 \;\;\;\;\;\;\;\;\;\;\mbox {(subtract second from first)}\\
y &= \frac{4}{19}x
\end{aligned}$
We can then find the $\displaystyle z$-coordinate of the line of intersection:
$\displaystyle z=x-2y=\frac{19}{19}x-2\cdot\frac{4}{19}x=\frac{11}{19}x.$
The equations of the line of intersection are therefore:
$\displaystyle \begin{aligned}
y&=\frac{4}{19}x\\
z&=\frac{11}{19}x.
\end{aligned}$
Now, all that remains is to find the value of $\displaystyle k$ that will result in these two equations of intersection for all three original equations.
You can check it by putting it into the original problem and seeing what happens!
If k= -7/19, then the equations become x-2y-z=0. x+9y-5z=0. and -(7/19)x-y+z=0 which we can rewrite as -7x- 19y+ 19z= 0. How many solutions does that have? If we subtract the first equation from the second, we eliminate x: (x+9y-5z)- (x-2y-z)= 11y- 4z= 0. If we multiply the first equation by 7 and add to the third, (7x-14y-7z)+(-7x-19y+19z)= -33y+ 12z= 0 or, dividing by 3, -11y+ 4z= 0, exactly the same equation as before! If z= (11/4)y ten the first equation becomes x- 2y- (11/4)y= 0 so x= (2+11/4)y= (33/4)y. That is any point on the line x= (33/4)t, y= t, z= (11/4)t lies on all three planes.