Explanation

(d)
Given that, $\overrightarrow{\mathbf{A}} = \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}\mathbf{,\ }\overrightarrow{\mathbf{B}} = \widehat{\mathbf{i}}\mathbf{,\ }\overrightarrow{\mathbf{C}} = c_{1}\widehat{\mathbf{i}} + c_{2}\widehat{\mathbf{j}} + c_{3}\widehat{\mathbf{k}}$
Since, $\overrightarrow{\mathbf{A}}$, $\overrightarrow{\mathbf{B}}$, $\overrightarrow{\mathbf{C}}$ are coplanar.
$\therefore\lbrack\overrightarrow{\mathbf{A}}$ $\overrightarrow{\mathbf{B}}$ $\overrightarrow{\mathbf{C}}$]=0
Now, $\overrightarrow{\mathbf{B}}$ $\mathbf{\times}\overrightarrow{\mathbf{C}} = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
1 & 0 & 0 \\
c_{1} & c_{2} & c_{3}
\end{matrix} \right| = - c_{3}\widehat{\mathbf{j}} + c_{2}\widehat{\mathbf{k}}$
$\therefore\ \overrightarrow{\mathbf{A}}\mathbf{\bullet (}\overrightarrow{\mathbf{B}}$ $\mathbf{\times}\overrightarrow{\mathbf{C}}\mathbf{) =}\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}} \right) \bullet \left( - c_{3}\widehat{\mathbf{j}} + c_{2}\widehat{\mathbf{k}} \right) = 0$
$\Rightarrow \$No value of $c_{1}$ can be found.