Let $A, B, C$ and $D$ are the position vectors of four points given as
$$
\begin{aligned}
& A(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), B(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}), C(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
& \text { and } D(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}})
\end{aligned}
$$
Here,
$$
\begin{aligned}
& \mathrm{AB}=-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\
& \mathrm{BC}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}
\end{aligned}
$$
and
$$
C D=5 \hat{i}+4 \hat{j}+(\lambda-2) \hat{k}
$$
As given that, these points are coplanar, therefore $[\mathrm{AB} \quad \mathrm{BC} \mathrm{CD}]=0$
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-1 & 5 & -3 \\
-3 & -2 & 6 \\
5 & 4 & \lambda-2
\end{array}\right|=0 \\
\Rightarrow \quad & -1(-2 \lambda+4-24)-5(-3(\lambda-2)-30) \\
\Rightarrow \quad & -3(-12+10)=0 \\
\Rightarrow \quad & 17 \lambda+20+20+15(\lambda-2)+150+36-30=0 \\
\Rightarrow \quad & 17 \lambda+146=0 \Rightarrow \lambda=-\frac{146}{17}
\end{aligned}
$$