According to the law of conservation of angular momentum, the angular momentum of a particle is conserved (constant) if the net external torque acting on it is zero . Mathematically, if $\vec{\tau} = 0$, then $\frac{d\vec{L}}{dt} = 0 \implies \vec{L} = \text{constant}$ . We know that torque $\vec{\tau} = \vec{r} \times \vec{F}$ . Given the position vector $\vec{r} = 2\hat{i} - 6\hat{j} - 12\hat{k}$ and the force vector $\vec{F} = \alpha\hat{i} + 3\hat{j} + 6\hat{k}$. Setting the torque to zero: $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix} = \vec{0}$ Expanding the determinant : $\hat{i}[(-6)(6) - (-12)(3)] - \hat{j}[(2)(6) - (-12)(\alpha)] + \hat{k}[(2)(3) - (-6)(\alpha)] = \vec{0}$ $\hat{i}[-36 + 36] - \hat{j}[12 + 12\alpha] + \hat{k}[6 + 6\alpha] = \vec{0}$ $0\hat{i} - (12 + 12\alpha)\hat{j} + (6 + 6\alpha)\hat{k} = 0\hat{i} + 0\hat{j} + 0\hat{k}$ Equating the coefficients of $\hat{j}$ and $\hat{k}$ to zero: $12 + 12\alpha = 0 \implies 12\alpha = -12 \implies \alpha = -1$ $6 + 6\alpha = 0 \implies 6\alpha = -6 \implies \alpha = -1$ Therefore, the value of $\alpha$ is $-1$.