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  4. The direction ratios of the normal to the plane passing through the points (1, -2, 3), (-1, 2, -1) and parallel to (x -2/2) = (y +1/3) = (z/4) is

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Q. The direction ratios of the normal to the plane passing through the points (1, -2, 3), (-1, 2, -1) and parallel to $\frac{x -2}{2} = \frac{y +1}{3} = \frac{z}{4}$ is

Three Dimensional Geometry

Solution:

Any plane through (1, -2, 3) is
A(x -1) + B(y + 2) + C(z - 3) = 0 ...(1)
The point (-1, 2, -1) lies in this plane if -2A + 4B- 4C = 0
i.e. if A - 2B+ 2C = 0 ...(2)
The plane (1) is parallel to the given line with d.r., 2, 3, 4
if 2A + 3B+ 4C = 0 ...(3)
From (2) and (3), we have
$\frac{A}{-8-6} = \frac{B}{4-4} = \frac{C}{3+4}$
$ \Rightarrow \frac{A}{-14} = \frac{B}{0} = \frac{C}{7} \Rightarrow A : B : C = 2 : 0 : - 1 $