Concept:

• The moment \(\rm \vec \tau\) of a force \(\rm \vec F\) about a line vector \(\rm \vec r\) is given by: \(\rm \vec \tau=\vec r \times\vec F\).
• A vector along the line with direction ratios l, m, n is simply the vector lî + mĵ + nk̂.

Calculation:

The direction ratios of the line of action of the force are 2, 2, 1.

So the force vector is given by:

\(\rm \vec F=78\frac{2\hat i+2\hat j+\hat k}{\sqrt{2^2+2^2+1^2}}\)

⇒ \(\rm \vec F=26\left(2\hat i+2\hat j+\hat k\right)\)

Now, the vector joining the point of action (2, 3, 5) with the origin is: 2î + 3ĵ + 5k̂.

The moment \(\rm \vec M\) of the force \(\rm \vec F\) about the origin = (2î + 3ĵ + 5k̂) × 26 (2î + 2ĵ + k̂) = 26(-7î + 8ĵ - 2k̂).

The moment of this force about the line joining the origin to the point (12, 3, 4) will be component of the above moment along this line.

The unit vector along this line is: \(\rm \frac{12\hat i+3\hat j+4\hat k}{\sqrt{12^2+3^2+4^2}}\) = \(\rm \frac{1}{13}\left(12\hat i+3\hat j+4\hat k\right)\).

The required magnitude of the component of the moment along the given line is: \(|\rm 26\left(-7\hat i+8\hat j-2\hat k\right).\frac{1}{13}\left(12\hat i+3\hat j+4\hat k\right)|\).

= |2(-84 + 24 - 12)|

= |2(-72)|

= 136.