Equation of the plane passing through 1,2,3 and containing the line r= 2i-j+k +u (i-2j-3k) . Please explain how to solve the question and the approach to it .

Since the plane is passing through pt. P(1,2,3) (hence OP= 1i + 2j + 3k  O being origin) and let pt,.A(2, –1,1) be the passing pt. of line hence PA = 1i –3j –2k which lies in the given plane and Hence PAxB = 5i + j +1k (B is the vector parallel to given line or D.R.s of line). Now this cross product is a vector normal to the plane.Hence equation of plane is r.n=OP.n (let n= PAxB) and cartesian equation is 5X+Y+Z=10.

please verify this answer and write comment how is this solution.