Question

Find an equation of a plane through the point (1, 5, 1) which is orthogonal to the line
x=3+5t y=5-1t z=-1+4t
in which the coefficient of x is 5.

Answer 1

The direction vector of the line is (5,-1,4)
so a plane perpendicular to that line is
5x - y + 4y + C = 0

but (1,5,1) is on it
5-5+1+C=0
C=-1

5x - y + 4y - 1 = 0