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Question

Find the equation of plane passing through the points P (1,1,1) , Q (3, -1, 2) and R (-3, 5 , -4).


A

6x + 16y +16z - 44 = 0

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B

x - 2y +16z - 13 = 0

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C

6x + 6y +16z - 28 = 0

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D

x + 6y +6z - 22 = 0

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Solution

The correct option is C

6x + 6y +16z - 28 = 0


Here one thing we need to understand is that question of finding the equations can be done in many ways. But it is always wise to choose those forms which can get us the answer easily. As in the end of the day it is not about solving it, it’s about solving it in the time limits.

The given question can be done with the help of general form, point normal form and by using the determinant form.

Out of all determinant form will be easiest. So we’ll use that.

∣ ∣xx1yy1zz1x2x1y2y1z2z1x3x1y3y1z3z1∣ ∣=0

Where (x,y,z) are the coordinates of any general point in the plane. And (x1,y1,z1),(x2,y2,z2),(x3,y3,z3) are the given points. Basically it’s nothing but the locus of any point (x,y,z) such that the volume of the tetrahedron made with the help of three fixed points is zero.

We’ll do the appropriate substitution.

∣ ∣x1y1z1311121315141∣ ∣=0

(x-1)(10 - 4) - (y -1)(-10 + 4) + (z -1)(8+8) = 0

6x + 6y +16z - 28 = 0


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