Question

If the lines $\frac{x-1}{2}=\frac{y}{-1}=\frac{z}{2}$ and $x-y+z-2=0=\lambda x+3z+5$ are coplanar, then the value of $7\lambda$ is equal to

• A. $31$
• B. $-52$
• C. $-39$
• D. $-31$

Answer 1

Answer: (D)

A plane passing through the line of intersection of the given $2$ planes is $\left(x-y+z-2\right)+\mu \left(\lambda x+3z+5\right)=0$
$\Rightarrow x\left(1+\lambda \mu \right)-y+z\left(1+3\mu \right)+5\mu -2=0$
Since the given line $\frac{x-1}{2}=\frac{y}{-1}=\frac{z}{2}$ must lie in this plane
$\Rightarrow 2\left(1+\lambda \mu \right)+\left(-1\right)\left(-1\right)+\left(1+3\mu \right)2=0$
$\Rightarrow 2+2\lambda \mu +1+2+6\mu =0$
$\Rightarrow 2\lambda \mu +6\mu +5=0\dots ..\left(i\right)$
The point $\left(1,0,0\right)$ must satisfy the equation of the plane
$\Rightarrow 1+\lambda \mu -0+0+5\mu -2=0$
$\Rightarrow \lambda \mu +5\mu -1=0\dots \dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$ we get,
$\mu =\frac{7}{4},\lambda =\frac{-31}{7}$
$\Rightarrow 7\lambda =-31$