b
For the conservation of angular momentum about origin, the torque \(\overrightarrow \tau  \) acting on the particle will be zero.

By definition, \(\overrightarrow \tau   = \overrightarrow r  \times F\)

Here, \(\overrightarrow r  = 2\hat i - 6\hat j - 12\hat k\,\, and\,\,  F  = \alpha \hat i + 3\hat j + 6\hat k\)

\(\begin{array}{l}
\therefore \,\,\,\,\overrightarrow \tau   = \left| \begin{array}{l}
\hat i\,\,\,\,\,\,\,\,\,\hat j\,\,\,\,\,\,\,\,\,\hat k\\
2\,\,\,\, - 6\,\,\,\, - 12\,\\
\alpha \,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,6
\end{array} \right|\\
\,\,\,\,\,\,\,\,\,\, = \hat i\left( { - 36 + 36} \right) - \hat j\left( {12 + 12\alpha } \right) + \hat k\left( {6 + 6\alpha } \right)\\
\,\,\,\,\,\,\,\,\,\, =  - \hat j\left( {12 + 12\alpha } \right) + \hat k\left( {6 + 6\alpha } \right)\\
But\,\overrightarrow \tau   = 0\\
\therefore \,\,\,\,\,12 + 12\alpha  = 0\,\,\,or\,\,\,\alpha  =  - 1\\
and\,\,6 + 6\alpha  = 0\,\,\,or\,\,\,\alpha  =  - 1\,\,
\end{array}\)