Answer
$x=4; y=3+6t; z=1+3t$
Work Step by Step
Let us take $z=1$
We have the equation of a plane: $x-2y+4=2$ and $x+y-2=5$ $\implies y=3$ and $x=7-y=7-3=4$
Thus, $r_0=\lt 4,3,1 \gt$
We know that $r=r_0+tv$ and $v=\lt 0,6,3 \gt$
Thus, our parametric equations become: $x=4+(0)t; y=3+6t; z=1+3t$
Hence, our parametric equations are: $x=4; y=3+6t; z=1+3t$