Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. The length of the projection of the vector $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

$A : (5,4,2)$

$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$

$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$

Any general point $P$ on the line is $(2\lambda - 1, 3\lambda + 3, -\lambda + 1)$

Let the given point is $A(5,4,2)$

$\overrightarrow{AP} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k}$

$\because \overrightarrow{AP} \perp \text{Line } (L)$

$\overrightarrow{AP} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$

$2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0$

$\Rightarrow \lambda = 1$

$\alpha = 1,; \beta = 6,; \gamma = 0$

Let the vector $\vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$

$\vec{u} = \hat{i} + 6\hat{j} + 0\hat{k}$

and $\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$

So projection $= \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|} = \frac{18}{7}$