Explanation

(d)
Let $\overrightarrow{\mathbf{c}} = 3\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 5\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{d}} = a\widehat{\dot{\mathbf{i}}} + b\widehat{\dot{\mathbf{j}}} - 15\widehat{\mathbf{k}}$
$For\ collinears,\ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
3 & 1 & - 5 \\
a & b & - 15
\end{matrix} \right| = \overrightarrow{\mathbf{0}}$
$\Longrightarrow \ \widehat{\dot{\mathbf{i}}}( - 15 + 5b) - \widehat{\dot{\mathbf{j}}}( - 45 + 5a) + \widehat{\mathbf{k}}(3b - a) = \overrightarrow{\mathbf{0}}$
$\Longrightarrow \ - 15 + 5b = 0,\ \ - 45 + 5a = 0,\ \ 3b - a = 0$
$\Longrightarrow \ \ b = 3,\ a = 9$