Explanation

(a)
Let $\overrightarrow{\mathbf{A}} = 2\widehat{\mathbf{i}} + 3\widehat{\mathbf{j}} - \widehat{\mathbf{k}}$ ….(i)
$\overrightarrow{\mathbf{B}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{+}2\widehat{\mathbf{j}}\mathbf{-}3\widehat{\mathbf{k}}$ ….(ii)
$\overrightarrow{\mathbf{C}} = 3\widehat{\mathbf{i}} + 4\widehat{\mathbf{j}} - 2\widehat{\mathbf{k}}$ …(iii)
$\overrightarrow{\mathbf{D}} = \widehat{\mathbf{i}} - \lambda\widehat{\mathbf{j}} + 6\widehat{\mathbf{k}}$ …..(iv)
From Eq. (i) and (ii), we get
$\overrightarrow{\mathbf{AB}}\mathbf{= -}\widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{j}}\mathbf{+}4\widehat{\mathbf{k}}$
$\therefore$ From Eq. (i) and (iii), we get
$\overrightarrow{\mathbf{AC}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{-}\widehat{\mathbf{k}}$
Similarly, from Eqs.(i) and (iv), we get
$\overrightarrow{\mathbf{AD}} = - \widehat{\mathbf{i}} - (\lambda - 3)\widehat{\mathbf{j}} + 7\widehat{\mathbf{k}}$
Now, using condition of coplanarity
$\left| \begin{matrix}
- 1 & - 1 & 4 \\
1 & 1 & - 1 \\
- 1 & - (\lambda + 3) & 7
\end{matrix} \right| = 0$
Applying $R_{1} \rightarrow R_{1} + R_{2}$, we get
$\left| \begin{matrix}
0 & 0 & 3 \\
1 & 1 & - 1 \\
- 1 & - (\lambda + 3) & 7
\end{matrix} \right| = 0$
$\Rightarrow \ - \lambda - 2 = 0\ \Rightarrow \ \lambda = - 2$