1.

Solve the question in the given attachment ​

Answer»

1) We have,

\begin{gathered}\vec{F}=ai+3j+6k\\ \\ \vec{r}=2i-6j-12k\end{gathered}

ANGULAR momentum about ORIGIN is conserved when torque about origin is 0.

That is,

\begin{gathered}\vec{r} \times \vec{F}=\vec{0}\\ \\= > \begin{vmatrix}i&j&k\\a&3&6\\2&-6&-12\end{vmatrix}=0\\ \\= > i*0+j(12+12a)+k(-6a-6)=0\\ \\= > 12+12a=0,-6a-6=0\\ \\= > \boxed{a=-1}\end{gathered}

Hence, VALUE of a is -1.


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