$t.(3,-2,3)$ and parallel $to$ the line $x+4=\frac{y}{2}=z-4$

The symmetric equations are given $by$

$x-3=\frac{y-2}{2}=z-3$

$\frac{x-3}{2}=y+2=\frac{z+3}{2}$

$x-3=\frac{y+2}{2}=z-3$

$\frac{x-3}{2}=y+2=z-3$

$x+3=\frac{y+2}{2}=z-3$