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Q.
Equation of line passing through the point $(2, 3, 1) $and parallel to the line of intersection of the plane $x- 2y - z + 5 = 0$ and $x + y + 3z = 6$ is
Let the DR’s of line passing through intersection of the given two planes $x - 2y - z + 5 = 0$ and $x + y + 3z = 6$ are $a, b$ and $c$.
Then, $ a - 2b - c = 0$
and $ a + b + 3c = 0 $
$\Rightarrow \frac{a}{-6+1}=\frac{-b}{3+1}=\frac{c}{1+2}$
$\Rightarrow a : b : c = -5 :-4 : 3$
Since, required line passes through (2, 3, 1) and parallel to above line. Hence, its equation will be
$ \frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}$