Find the equation of the plane through (1,2,-2) that contains the line x = 2t, y = 3 - t, z = 1 +3t

Answer 1

The direction vector of the given line is (2,-1,3) and a point on it is (0,3,1)
So using that point and the additional given point (1,2,-2)
yields a second direction vector of (1,-1,-3)

So we need a normal to these two direction vectors, which is the cross-product of (2,-1,3) and (1,-1,-3)

I got this normal to be (6,9,-1)
(I will assume you know how to find the cross-product)

The equation of the plane is
6x + 9y - z = k
subbing in the given point (1,2,-2)
6 + 18 + 2 = k = 26

equation of plane : 6x + 9y - z = 26