Let $$L_1:\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-1}{2}$$ and $$L_2:\frac{x-2}{1}=\frac{y-3}{4}=\frac{z}{1}$$ be two lines in space. $$M$$ and $$N$$ are points on $$L_1$$ and $$L_2$$ respectively such that $$MN$$ is the shortest distance between $$L_1$$ and $$L_2$$. What is the sum of the coordinates of $$M$$?

Any general point $$M$$ on line $$L_1$$ can be parameterized as:

$$M(3\lambda + 1, \lambda + 2, 2\lambda + 1)$$

Any general point $$N$$ on line $$L_2$$ can be parameterized as:

$$N(\mu + 2, 4\mu + 3, \mu)$$

The direction ratios of the vector $$\vec{MN}$$ are given by:

$$\vec{MN} = (\mu - 3\lambda + 1, 4\mu - \lambda + 1, \mu - 2\lambda - 1)$$

Since $$\vec{MN}$$ is the shortest distance line, it must be perpendicular to the direction vectors of both $$L_1$$ and $$L_2$$.

The direction vectors are $$\vec{b_1} = (3, 1, 2)$$ and $$\vec{b_2} = (1, 4, 1)$$.

Taking the dot product of $$\vec{MN}$$ with $$\vec{b_1}$$:

$$3(\mu - 3\lambda + 1) + 1(4\mu - \lambda + 1) + 2(\mu - 2\lambda - 1) = 0$$

$$3\mu - 9\lambda + 3 + 4\mu - \lambda + 1 + 2\mu - 4\lambda - 2 = 0$$

$$9\mu - 14\lambda + 2 = 0 \quad \text{--- (Equation 1)}$$

Taking the dot product of $$\vec{MN}$ with $\vec{b_2}$$:

$$1(\mu - 3\lambda + 1) + 4(4\mu - \lambda + 1) + 1(\mu - 2\lambda - 1) = 0$$

$$\mu - 3\lambda + 1 + 16\mu - 4\lambda + 4 + \mu - 2\lambda - 1 = 0$$

$$18\mu - 9\lambda + 4 = 0 \quad \text{--- (Equation 2)}$$

Solving Equations 1 and 2 simultaneously:

From Equation 1, $$9\mu = 14\lambda - 2$$.

Substituting into Equation 2:

$$2(14\lambda - 2) - 9\lambda + 4 = 0$$

$$28\lambda - 4 - 9\lambda + 4 = 0 \implies 19\lambda = 0 \implies \lambda = 0$$

Substituting $$\lambda = 0$$ into the coordinates of point $$M$$:

$$M(3(0) + 1, 0 + 2, 2(0) + 1) = (1, 2, 1)$$

The sum of the coordinates of point $$M$$ is:

$$1 + 2 + 1 = 4$$

The correct option is C.