Let $$ L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \text{ and } L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5} $$ be two lines. Then which of the following points lies on the line of the shortest distance between $$L_1 \text{ and } L_2 $$ ?

Given lines $$L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ and $$L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$.

Point on $$L_1$$: $$A = (1, 2, 3)$$, direction $$\vec{d_1} = (2, 3, 4)$$.

Point on $$L_2$$: $$B = (2, 4, 5)$$, direction $$\vec{d_2} = (3, 4, 5)$$.

The shortest distance line is perpendicular to both $$L_1$$ and $$L_2$$, so its direction is $$\vec{d_1} \times \vec{d_2}$$:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = (15-16)\hat{i} - (10-12)\hat{j} + (8-9)\hat{k} = (-1, 2, -1)$$

Let the point on $$L_1$$ be $$P = (1+2t, 2+3t, 3+4t)$$ and on $$L_2$$ be $$Q = (2+3s, 4+4s, 5+5s)$$.

$$\vec{PQ}$$ must be parallel to $$(-1, 2, -1)$$:

$$\vec{PQ} = (1+3s-2t, 2+4s-3t, 2+5s-4t)$$

Setting $$\vec{PQ} \cdot \vec{d_1} = 0$$:

$$2(1+3s-2t) + 3(2+4s-3t) + 4(2+5s-4t) = 0$$

$$2+6s-4t+6+12s-9t+8+20s-16t = 0$$

$$16 + 38s - 29t = 0 \quad \cdots (1)$$

Setting $$\vec{PQ} \cdot \vec{d_2} = 0$$:

$$3(1+3s-2t) + 4(2+4s-3t) + 5(2+5s-4t) = 0$$

$$3+9s-6t+8+16s-12t+10+25s-20t = 0$$

$$21 + 50s - 38t = 0 \quad \cdots (2)$$

From (1): $$29t = 16 + 38s \implies t = \frac{16+38s}{29}$$

Substituting into (2): $$21 + 50s - 38 \cdot \frac{16+38s}{29} = 0$$

$$29(21+50s) - 38(16+38s) = 0$$

$$609 + 1450s - 608 - 1444s = 0$$

$$1 + 6s = 0 \implies s = -\frac{1}{6}$$

$$t = \frac{16 + 38(-1/6)}{29} = \frac{16 - 19/3}{29} = \frac{29/3}{29} = \frac{1}{3}$$

$$P = \left(1+\frac{2}{3},\; 2+1,\; 3+\frac{4}{3}\right) = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$$

$$Q = \left(2-\frac{1}{2},\; 4-\frac{2}{3},\; 5-\frac{5}{6}\right) = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$$

The line of shortest distance passes through $$P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$$ with direction $$(-1, 2, -1)$$:

$$\text{Parametric form: } \left(\frac{5}{3} - \lambda,\; 3 + 2\lambda,\; \frac{13}{3} - \lambda\right)$$

Checking Option (1): $$\left(\frac{14}{3}, -3, \frac{22}{3}\right)$$:

$$\frac{5}{3} - \lambda = \frac{14}{3} \implies \lambda = -3$$

$$3 + 2(-3) = 3 - 6 = -3 \quad \checkmark$$

$$\frac{13}{3} - (-3) = \frac{13}{3} + 3 = \frac{22}{3} \quad \checkmark$$

The correct answer is Option (1): $$\boxed{\left(\frac{14}{3}, -3, \frac{22}{3}\right)}$$.