## Thomas' Calculus 13th Edition

$x=2-2t,y=3+4t, z=-2t$
As we know the parametric equations of a straight line for a vector $v=v_1i+v_2j+v_3k$ passing through a point $P(a,b,c)$ is given by $x=a+t v_1,y=b+t v_2; z=c+t v_3$ Consider two vectors u and v and line is perpendicular to these vectors.This implies that these vectors are parallel to the cross product of $n=u \times v$ and $n=-2i+4j-2k$ So, we have $v=\lt -2,4,-2 \gt$ and $P=(2,3,0)$ . Our parametric equations are: $x=2-2t,y=3+4t, z=-2t$