Answer :

Given that,

The line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3).


We need to find the equation of the plane.


We must understand that,


Any line which is perpendicular to the plane is the normal.


Let the points be P(-2, -1, -3) and Q(1, -3, 3), so the line PQ is perpendicular to the plane.


This also means, PQ is normal to the plane.


Therefore, PQ = (1 – (-2), -3 – (-1), 3 – (-3))


PQ = (1 + 2, -3 + 1, 3 + 3)


PQ = (3, -2, 6)


Normal to the plane =



The vector equation of a plane is given by



Put these values in the above equation,





We get






3(x – 1) + (-2)(y + 3) + 6(z – 3) = 0


3(x – 1) – 2(y + 3) + 6(z – 3) = 0


3x – 3 – 2y – 6 + 6z – 18 = 0


3x – 2y + 6z – 3 – 6 – 18 = 0


3x – 2y + 6z – 9 – 18 = 0


3x – 2y + 6z – 27 = 0


3x – 2y + 6z = 27


Thus, required equation of the plane is 3x – 2y + 6z = 27.


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