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 (x_{1},y_{1},z_{1}) and perpendicular to a line with direction ratios A, B, C is

A(x - x_{1}) + B(y - y_{1}) + C(z - z_{1}) = 0

The plane passes through P(1,2,3)

So, x_{1} = 1, y_{1} = 2, z_{1} = - 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

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