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(C) The equation of the line is $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$.This can be written in Cartesian

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$\frac{18}{7}$

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(C) The equation of the line is $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$.This can be written in Cartesian form as $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$.Any general point $P$ on the line is given by $(2\lambda - 1, 3\lambda + 3, -\lambda + 1)$.Let the given point be $A = (5, 4, 2)$.The vector $\vec{AP} = (2\lambda - 1 - 5)\hat{i} + (3\lambda + 3 - 4)\hat{j} + (-\lambda + 1 - 2)\hat{k} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k}$.Since $\vec{AP}$ is perpendicular to the line, its dot product with the direction vector of the line $(2\hat{i} + 3\hat{j} - \hat{k})$ must be zero.$(2\lambda - 6)(2) + (3\lambda - 1)(3) + (-\lambda - 1)(-1) = 0$.$4\lambda - 12 + 9\lambda - 3 + \lambda + 1 = 0$.$14\lambda - 14 = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ into the coordinates of $P$, we get $\alpha = 2(1) - 1 = 1$, $\beta = 3(1) + 3 = 6$, and $\gamma = -1 + 1 = 0$.So, the vector $\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} = \hat{i} + 6\hat{j} + 0\hat{k}$.Let $\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$.The length of the projection of $\vec{u}$ on $\vec{w}$ is given by $\frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|}$.$\vec{u} \cdot \vec{w} = (1)(6) + (6)(2) + (0)(3) = 6 + 12 + 0 = 18$.$|\vec{w}| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.Therefore, the projection length is $\frac{18}{7}$.

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