If O be the origin and the coordinates of P be (1, 2, –3), then find the equation of the plane passing through P and perpendicular to OP.
The equation of a plane passing through (x1,y1,z1) and perpendicular to a line with direction ratios A, B, C is
A(x - x1) + B(y - y1) + C(z - z1) = 0
The plane passes through P(1,2,3)
So, x1 = 1, y1 = 2, z1 = - 3
Normal vector to plane = 
where O(0,0,0), P(1,2, - 3)
Direction ratios of 
= (1 - 0), (2 - 0), (-3 - 0)
= (1,2, - 3)
∴ A = 1, B = 2, C = -3
Equation of plane in cartesian form is
1(x - 1) + 2(y - 2) - 3(z - (-3)) = 0
⇒ x - 1 + 2y - 4 - 3z - 9 = 0
⇒ x + 2y - 3z - 14 = 0
16