Posted by Phil on Thursday, February 17, 2011 at 3:12pm.
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