Answer :

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


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