intersection vectors problem

Question

Find the intersection of xyz=[1,4,−2]+t[1,−1,3] and 2x−5y+8z−28=0.

Answer 1

The graph of (x,y,z) = (1, 4, -2) + t(1, -1, 3) is a line and the graph of 2x - 5y + 8z - 28 = 0 is a plane.

x = 1 + t

y = 4 - t

z = -2 + 3t

So, 2(1 + t) - 5(4 - t) + 8(-2 + 3t) - 28 = 0

31t -62 = 0 So, t = 2.

x = 3, y = 2, and z = 4

The point of intersection of the line and the plane is (3, 2, 4)