Question

$${\overline{r} = ( - \widehat{i} + 3\widehat{j} + \widehat{k}) + \lambda(2\widehat{i} + 3\widehat{j} - \widehat{k}) }{\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} = \lambda }$$Any general point P on the line is

$$(2\lambda - 1,3\lambda + 3, - \lambda + 1) $$Let the given point is $A(5,4,2)$

$$\overline{AP}(2\lambda - 6)\widehat{i} + (3\lambda - 1)\widehat{j} + ( - \lambda - 1)\widehat{k} $$$\because\overline{AP}\bot^{r}$ Line $(L)$

$${\therefore\overline{AP} \cdot (2\widehat{i} + 3\widehat{j} - \widehat{k}) = 0 }{2(2\lambda - 6) + 3(3\lambda - 1) - 1( - \lambda - 1) = 0 }{\Rightarrow \lambda = 1 }{\therefore\alpha = 1 }{\beta = 6 }{\gamma = 0 }$$Let the vector $\overline{u} = \alpha\widehat{i} + \beta\widehat{j} + \gamma\widehat{k}$

$$\overline{u} = \widehat{i} + 6\widehat{j} + O\widehat{k}$$

$$\&\overline{w} = 6\widehat{i} + 2\widehat{j} + 3\widehat{k} $$So projection $= \frac{|\overline{u} \cdot \overline{w}|}{|\overline{w}|} = \frac{18}{7}$