## Calculus 8th Edition

$x=-6+2t,y=2+3t,z=3+t$ and $\frac{x+6}{2}=\frac{y-2}{3}=z-3$
The vector for is through the $(-6,2,3)$ 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: $\frac{x+6}{2}=\frac{y-2}{3}=z-3$