The moment about a line through the origin having the direction of $2\hat i-2\hat j+\hat k$ due to a 30 kg force acting at a point (-4, 2, 5) in the direction of $12\hat i -4\hat j-3\hat k$, is

$\frac{-760}{13}$

Let $\vec F$ be the force. Then,

$\vec F=\frac{30(12\hat i -4\hat j-3\hat k)}{\sqrt{12^2+(-4)^2+(-3)^2}}=\frac{30}{13}(12\hat i -4\hat j-3\hat k)$

Suppose the force $\vec F$ acts at point P (-4, 2, 5). The moment of $\vec F$ acting at P about a line in the direction $2\hat i -2\hat j+\hat k$ is equal to the resolve part along the line of the moment of $\vec F$ about a point on the line. It is given that the line passes through the origin O.

So we choose O as a point on the line. Let $\vec{OP} =\vec r$. Then

$⇒\vec r =\vec{OP}$ = Position vector of P - Position vector of O

$⇒\vec r =(-4\hat i+2\hat j+5\hat k)-\vec 0=-4\hat i+2\hat j+5\hat k$

Let $\vec M$ be the moment of $\vec F$ about O. Then,

$\vec M=\vec r×\vec F$

$⇒\vec M=(-4\hat i+2\hat j+5\hat k) ×\frac{30}{13} (12\hat i -4\hat j-3\hat k)$

$⇒\vec M=\frac{30}{13}\begin{vmatrix}\hat i&\hat j&\hat k\\-4&2&5\\12&-4&-3\end{vmatrix}=\frac{30}{13}(14\hat i+48\hat j-8\hat k)$

Let $\vec a$ be unit vector in the direction of $2\hat i -2\hat j+\hat k$. Then,

$\vec a=\frac{2\hat i-2\hat j+\hat k}{\sqrt{4+4+1}}=\frac{1}{3}(2\hat i-2\hat j+\hat k)$

Thus, the moment of $\vec F$ about the given line

$=\vec M.\vec a=\frac{30}{13}(14\hat i+48\hat j-8\hat k).\frac{1}{3}(2\hat i-2\hat j+\hat k)=\frac{-760}{13}$