Question

Find the point at which the line with parametric equations x = 2 3t, y =-4t, z =5 t intersects the plane 4x 5y -2z=18

Answer 1

It looks like you have

$\begin{cases}x=2+3t\\y=-4t\\z=5+t\end{cases}$

Substituting these in to the plane equation, we have

$4x+5y-2z = 4(2+3t) + 5(-4t) - 2(5+t) = -2-10t = 18 \\\\ \implies-10t = 20 \implies t = -2$

When $t=-2$, we get the point

$\begin{cases}x=2+3(-2)\\y=-4(-2)\\z=5+(-2)\end{cases} \implies (x,y,z) = \boxed{(-4,8,3)}$