If the straight lines $x = 1+s, y = -3 - λs, z = 1 + λs$ and $x=\frac{t}{2}, y = 1 + t, z = 2-t$ with parameters s and t respectively, are coplanar, then λ equals

-2

Given lines are

$\frac{x-1}{1}=\frac{y+3}{-λ}=\frac{z-1}{λ}= s$ and, $\frac{x-0}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}= t $

We know that the lines

$\frac{x-x_1}{1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and, $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$

are coplanar, if

$\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1\\l_1 & m_1 & n_1\\l_2 & m_2 & n_2\end{vmatrix}=0$

So, the given lines will be coplanar, if

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