find the direction and one point on the following line
$\displaystyle x + 1 = -\frac{y-5}{3} = \frac{3-z}{2} $
am not sure how to do this ?
I have simplified the equation and get
$\displaystyle 6x+2y+3z-13 =0 $
The result of your simplification is a single equation, which defines a plane in 3D space (two variables are free and one depends on them). The original formula consists of two first-order equations. Each of them determines a plane, and together they determine the intersection of those planes, i.e., a line.
Suppose the following equations are given.
$$
\frac{x-x_0}{A}=\frac{y-y_0}{B} =\frac{z-z_0}{C}
$$
If we denote the value of these three fractions by $t$ and express $x,y,z$ through $t$, we get a system of parametric equations
$$
\begin{aligned}
x&=x_0+At\\
y&=y_0+Bt\\
z&=z_0+Ct
\end{aligned}
$$
They determine a line passing through $(x_0,y_0,z_0)$ in the direction $(A,B,C)$.
The way you have rewritten it is completely wrong. You have turned a line into a plane.
This is the symmetric form $\dfrac{x-a}{n_1}=\dfrac{y-b}{n_2}=\dfrac{z-c}{n_3}$ of a line.
The point $(a,b,c)$ is on that line, and the vector $\vec{D}=<n_1,n_2,n_3>$ is the direction vector.
The vector form is $<a,b,c>+t\cdot \vec{D}$. The parametric form is $\begin{array}{*{20}{c}}
{x = a + t{n_1}} \\
{y = b + t{n_2}} \\
{z = c + t{n_3}} \end{array}$
Write this as $\dfrac{x + 1}{1} = \dfrac{y-5}{-3} = \dfrac{z-3}{-2}$ now we have a point $(-1,5,3)$ and direction vector $<1,-3,-2>$