Question

Equation of the line through the point $(2,3,1)$ and parallel to the line of intersection of the planes $x-2y-z+5=0$ 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. $\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{3}$
• D. $\frac{x-2}{4}=\frac{y-3}{3}=\frac{z-1}{2}$
• E. $\frac{x-2}{-4}=\frac{y-3}{-3}=\frac{z-1}{2}$

Answer 1

The correct option is A $\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}$

Given equation of planes are
$x-2y-z+5=0$
and $x+y+3z=6$.
Firstly, determine the intersection lines of two planes.
Let the DR's of intersection line are $a,b$ and $c$.
Since, the normal to the given planes are perpendicular to the intersecting line.
$\therefore a\left(1\right)+b(-2)+c(-1)=0$
$\Rightarrow a-2b-c=0$ ..... (i)
and $a\left(1\right)+b\left(1\right)+c\left(3\right)=0$
$\Rightarrow a+b+3c=0$ ...... (ii)
From Eqs. (i) and (ii), we get
$\frac{a}{-6+1}=\frac{b}{-1-3}=\frac{c}{1+2}$
$\Rightarrow \frac{a}{-5}=\frac{b}{-4}=\frac{c}{3}$
Since, the required line is passing through $(2,3,1)$ and parallel to the line of intersection.
Therefore, $\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}$