Answer
$x=-2+2t,y=2-t,z=4+5t$
Work Step by Step
Given: The line $(-2,2,4)$ perpendicular to $2x-y+5z=12$
The direction of the vectors is $<2,-1,5>$
Formula for a line$ r=r_0+tv $
where$ r_0$ is the starting point vector and $v$ is the direction vector.
$r=+t<2,-1,5> $
Therefore, the parametric equations of the line are:
$x=-2+2t,y=2-t,z=4+5t$
