Find the equation of the plane which passes through the point P( 2,1,3) and contains the line
(x + 1)/3 = (y-1 )/2= 5/z
I assume you meant z/5, not 5/z
Let P = (-1,1,0) and Q=(2,1,3) and v=<3,2,5>
Then PQ = P-Q = (-3,0,-3) and the normal vector is PQ×v = <-6,24,-6>
and the plane we want is thus -6(x-2)+24(y-1)-6(z-3) = 0
or, x-4y+z = 1