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

Foot of perpendicular from (5,4,2) on line $$\vec{r} = (-1,3,1) + \lambda(2,3,-1)$$.

Point on line: $$(-1+2\lambda, 3+3\lambda, 1-\lambda)$$.

Direction from (5,4,2) to point: $$(-6+2\lambda, -1+3\lambda, -1-\lambda)$$.

Perpendicular to direction (2,3,-1):

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

$$-12+4\lambda-3+9\lambda+1+\lambda = 0$$

$$14\lambda - 14 = 0 \Rightarrow \lambda = 1$$

Foot = $$(1, 6, 0)$$, so $$(\alpha, \beta, \gamma) = (1, 6, 0)$$.

Projection of $$(1, 6, 0)$$ on $$(6, 2, 3)$$: $$\frac{6+12+0}{\sqrt{36+4+9}} = \frac{18}{7}$$.

The answer is Option 4: $$\frac{18}{7}$$.