A force $\overrightarrow{\mathrm{F}}=\mathrm{\alpha }\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}$ is acting at a point $\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}-12\widehat{\mathrm{k}}$. The value of $\mathrm{\alpha }$ for which angular momentum about the origin is conserved is

Given, 

$\overrightarrow{\mathrm{F}}=\mathrm{\alpha }\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}-12\widehat{\mathrm{k}}$

We know,

$\overrightarrow{\mathrm{\tau }}=\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{F}}$

$\therefore \overrightarrow{\mathrm{\tau }}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& -6& -12\\ \mathrm{\alpha }& 3& 6\end{array}\right|$

$=\hat{\mathrm{i}}(-36+36)-\hat{\mathrm{j}}(12+12\mathrm{\alpha })+\hat{\mathrm{k}}(6+6\mathrm{\alpha })$

$=\hat{\mathrm{i}}\left(0\right)-\hat{\mathrm{j}}(12+12\mathrm{\alpha })+\hat{\mathrm{k}}(6+6\mathrm{\alpha })$

For the conservation of angular momentum about the origin, the torque t acting on the particle will be zero.

$\therefore 12+12\mathrm{\alpha }=0 \mathrm{and} 6+6\mathrm{\alpha }=0$

$\Rightarrow \mathrm{\alpha }=-1$