Answer
$x=-6+2t,y=2+3t,z=3+t$
and
$\frac{x+6}{2}=\frac{y-2}{3}=z-3$
Work Step by Step
The vector is through the point $(-6,2,3)$ and is parallel to $1/2x=1/3y=z+1$
Equation of the line is given by $r=r_0+tv$
$r_0=(-6,2,3)$ and $v= \lt 2,3,1 \gt$
Parametric equations are defined by:
$x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$
Thus, the parametric equations are:
$x=-6+2t,y=2+3t,z=3+t$
The symmetric equations are defined by:
$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$
Hence, the symmetric equations are:
$\frac{x+6}{2}=\frac{y-2}{3}=z-3$
