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 $\left(-1,0,1\right)$
$\overrightarrow{\mathrm{OA}}=-\hat{i}+\hat{k}$
$\because$ A normal vector to required plane is $\overrightarrow{\mathrm{OA}}$
Equation of the required plane is $-1\left(x-0\right)+1\left(z-0\right)=0$
$\Rightarrow -x+z=0$
$\Rightarrow 10x-10z=0$
$b=0,c=-10,d=0$