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 co-planar then $$\lambda$$ equals

For the first line,

$$A=(1,-3,1)$$

and its direction vector is

$$\vec d_1=(1,-\lambda,\lambda)$$

For the second line,

$$B=(0,1,2)$$

and its direction vector is

$$\vec d_2=\left(\frac12,1,-1\right)$$

The vector joining the points $$A$$ and $$B$$ is

$$\overrightarrow{AB}=(-1,4,1)$$

For two lines to be coplanar,

$$[\overrightarrow{AB},\vec d_1,\vec d_2]=0$$

Therefore,
$$\begin{vmatrix}-1 & 4 & 1\\1 & -\lambda & \lambda\\\frac12 & 1 & -1\end{vmatrix}=0$$

Expanding,
$$(-1)\begin{vmatrix}-\lambda & \lambda\\1 & -1\end{vmatrix}-4\begin{vmatrix}1 & \lambda\\\frac12 & -1\end{vmatrix}+\begin{vmatrix}1 & -\lambda\\\frac12 & 1\end{vmatrix}=0$$

$$(-1)(\lambda-\lambda)-4\left(-1-\frac{\lambda}{2}\right)+\left(1+\frac{\lambda}{2}\right)=0$$

$$4+2\lambda+1+\frac{\lambda}{2}=0$$

$$5+\frac{5\lambda}{2}=0$$

$$10+5\lambda=0$$

$$\lambda=-2$$

Hence,
$$\boxed{-2}$$