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

The given straight line is

$\begin{array}{l}\mathrm{x}=1+\mathrm{s}, \mathrm{y}=-3-\mathrm{\lambda s}, \mathrm{z}=1+\mathrm{\lambda s}\\ \therefore \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3}{-\mathrm{\lambda }}=\frac{\mathrm{z}-1}{\mathrm{\lambda }}=\mathrm{s}\end{array}$

Also the given equation of another straight line is

$\mathrm{x}=\frac{\mathrm{t}}{2}, \mathrm{y}=1+\mathrm{t}, \mathrm{z}=2-\mathrm{t}\therefore \frac{\mathrm{x}-0}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{-2}=\mathrm{t}$

These two lines are coplanar if

$\Rightarrow \left|\begin{array}{ccc}1& -4& -1\\ 1& -\mathrm{\lambda }& \mathrm{\lambda }\\ 1& 2& -2\end{array}\right|=0$