Explanation

(c)
Let $\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}},\ \ \ \overrightarrow{\mathbf{b}} = 2\widehat{\dot{\mathbf{i}}} + 0\widehat{\dot{\mathbf{j}}} + 0\widehat{\mathbf{k}}$
Now take option (c).
Let $\overrightarrow{\mathbf{c}} = 0\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} - 6\widehat{\mathbf{k}}$
$Now,\ \ \overrightarrow{\mathbf{a}} \bullet \left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} \right) = \left| \begin{matrix}
1 & - 2 & - 3 \\
2 & 0 & 0 \\
0 & - 4 & - 6
\end{matrix} \right|$
$= 1(0) + 2( - 12) - 3( - 8) = 0$