The position vectors of the points $A, B, C$ and $D$ are $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and
$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\lambda \hat{\mathbf{k}}$ respectively
$\begin{aligned}
\mathbf{B A} &=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\mathbf{k})-(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \mathbf{k})=\hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \mathbf{k} \\
\mathbf{C A} &=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \mathbf{k})=-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
\mathbf{D A} &=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}) \\
&=-\hat{\mathbf{i}}-\hat{\mathbf{j}}-(1+\lambda) \mathbf{k}
\end{aligned}$
These are coplanar, so
$\left|\begin{array}{ccc}1 & 1 & -3 \\ -2 & -1 & -3 \\ -1 & -1 & -(1+\lambda)\end{array}\right|=0$
$=\left|\begin{array}{ccc}-1 & -1 & 3 \\ 2 & 1 & 3 \\ 1 & 1 & 1+\lambda\end{array}\right|=0$
$-1[1+\lambda-3]+1[2+2 \lambda-3]+3[2-1]=0$
$\Rightarrow \quad-1[\lambda-2]+1[2 \lambda-1]+3=0$
$\Rightarrow -\lambda+2+2 \lambda-1+3=0$
$\Rightarrow \lambda+4=0$
$\Rightarrow \lambda=-4$