Question

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

• A. $\frac{x-2}{5}=\frac{y-3}{-4}=\frac{z-1}{3}$
• B. $\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}$
• C. $\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{2}$
• D. $\frac{x-2}{4}=\frac{y-3}{4}=\frac{z-1}{2}$

Answer 1

Answer: (B)

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
$\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}$