Answer
$x=６-2t, y=-2+3t, z=-3+t$
Work Step by Step
Step 1. Establish a vector along the line $\vec v=\vec {PQ}=\langle 4-6,1+2,-2+3 \rangle=\langle -2,3,1 \rangle$
Step 2. The general parametric equations for a line passing point $P(x_0,y_0,z_0)$ and parallel to a vector $\langle a,b,c \rangle$ are $x=x_0+at, y=y_0+bt, z=z_0+ct$
Step 3. We can obtain the equations for this problem as: $x=６-2t, y=-2+3t, z=-3+t$
