Question

Let $L_{1}:x=y=z$ and $L_{2}:x-1=y-2=z-3$ be two lines. The foot of perpendicular drawn from the origin $O\left(0,0 , 0\right)$ on $L_{1}$ to $L_{2}$ is $A.$ If the equation of a plane containing the line $L_{1}$ and perpendicular to $OA$ is $10x+by+cz=d$ , then the value of $b+c+d$ is equal to

• A. $10$
• B. $-10$
• C. $12$
• D. $-7$

Answer 1

Answer: (B)

From diagram,
$OA$ is perpendicular to line $L_{2}=0$
$\Rightarrow 1\left(\lambda + 1\right)+1\left(\lambda + 2\right)+1\left(\lambda + 3\right)=0\Rightarrow \lambda =-2$
So, coordinates of point $A$ are $\left(- 1,0 , 1\right)$
$\overset{ \rightarrow }{O A}=-\hat{i}+\hat{k}$
$\because $ A normal vector to required plane is $\overset{ \rightarrow }{O A}$
Equation of the required plane is $-1\left(x - 0\right)+1\left(z - 0\right)=0$
$\Rightarrow -x+z=0$
$\Rightarrow 10x-10z=0$
$b=0,c=-10,d=0$