A force of 78 grams acts at the point (2,3,5). The direction ratios of the line of action being 2,2,1 . The magnitude of its moment about the line joining the origin to the point (12,3,4) is

Given: Force magnitude \(=78\) (grams), point of application \(\mathbf{r}=(2,3,5)\), force direction ratios \((2,2,1)\). Line (about which moment is required) is along \((12,3,4)\) through the origin.

Force vector: Unit along \((2,2,1)\) is \(\dfrac{1}{\sqrt{2^2+2^2+1^2}}(2,2,1)=\dfrac{1}{3}(2,2,1)\). Hence \[ \mathbf{F}=78\cdot \frac{1}{3}(2,2,1)=(52,52,26). \]

Torque about origin: \[ \boldsymbol{\tau}=\mathbf{r}\times \mathbf{F}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&3&5\\ 52&52&26 \end{vmatrix} =(-182,\,208,\,-52). \]

Moment about the line with direction \(\mathbf{L}=(12,3,4)\): Let \(\hat{\ell}=\dfrac{\mathbf{L}}{\lVert \mathbf{L}\rVert}\). The required magnitude is the component of \(\boldsymbol{\tau}\) along \(\hat{\ell}\): \[ M=\big|\boldsymbol{\tau}\cdot \hat{\ell}\big|=\frac{\big|\boldsymbol{\tau}\cdot \mathbf{L}\big|}{\lVert \mathbf{L}\rVert} =\frac{|(-182,208,-52)\cdot(12,3,4)|}{\sqrt{12^2+3^2+4^2}} =\frac{|{-1768}|}{13}=136. \]

Answer: \( \boxed{136} \) (in gram–unit of length).