Calculus
posted by Phil .
Find the equation of the plane through (1,2,2) that contains the line x = 2t, y = 3  t, z = 1 +3t

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 crossproduct 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 crossproduct)
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