Explanation

(c)
Let $\overrightarrow{\mathbf{A}} = 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}},$ $\overrightarrow{\mathbf{B}} = \widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$,
$\overrightarrow{\mathbf{C}} = 3\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{D}} = \widehat{\dot{\mathbf{i}}} - 6\widehat{\dot{\mathbf{j}}} + \lambda\widehat{\mathbf{k}}$
Now, $\overrightarrow{\mathbf{AB}} = - \widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}},\ \overrightarrow{\mathbf{A}}C = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
$and\ \overrightarrow{\mathbf{AD}} = - \widehat{\dot{\mathbf{i}}} - 9\widehat{\dot{\mathbf{j}}} + (\lambda + 1)\widehat{\mathbf{k}}$
These will be coplanar, if $\left\lbrack \overrightarrow{\mathbf{AB}}\ \overrightarrow{\mathbf{AC}}\ \overrightarrow{\mathbf{AD}} \right\rbrack = 0$
$\therefore\ \left| \begin{matrix}
- 1 & - 5 & 4 \\
1 & 1 & - 1 \\
- 1 & - 9 & (\lambda + 1)
\end{matrix} \right| = 0$
$\Longrightarrow \ - 1(\lambda + 1 - 9) + 5(\lambda + 1 - 1) + 4( - 9 + 1) = 0$
$\Longrightarrow \ \lambda = 6$