If the straight lines $x=1+s, y=-3-\lambda s, z=1+\lambda s$ and $x=\frac{t}{2}, y=1+t, z=2-t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to:

-2

The correct answer is Option (2) → -2

$L_1:\frac{x-1}{1}=\frac{y+3}{-λ}=\frac{z-1}{λ}$

$L_2:\frac{x}{\frac{1}{2}}=\frac{y-1}{1}=\frac{z-2}{-1}$

so $Δ=\begin{vmatrix}0-1&1-(-3)&2-1\\1&-λ&λ\\\frac{1}{2}&1&-1\end{vmatrix}=0$

$⇒Δ=\begin{vmatrix}-1&4&1\\1&-λ&λ\\\frac{1}{2}&1&-1\end{vmatrix}=0$

$C_3→C_3+C_2$

$Δ=\begin{vmatrix}-1&4&5\\1&-λ&0\\\frac{1}{2}&1&0\end{vmatrix}=0$

$⇒5(1+\frac{λ}{2})=0⇒λ=-2$