Equation of the plane which passes through the point of intersection of lines $$\frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2}$$ and $$\frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3}$$ and has the largest distance from the origin is:

First Line: $$(x, y, z) = (3\lambda + 1, \lambda + 2, 2\lambda + 3)$$

Second line: $$(x, y, z) = (\mu + 3, 2\mu + 1, 3\mu + 2)$$

Point of intersection:  $$3\lambda + 1 = \mu + 3 \implies \mu = 3\lambda - 2$$

$$\lambda + 2 = 2\mu + 1$$

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

$$5\lambda = 5 \implies \lambda = 1$$

Thus, the point of intersection is $$P(4, 3, 5)$$

For a plane passing through a fixed point $$P$$ to have the maximum possible perpendicular distance from the origin $$O(0,0,0)$$, the line segment $$OP$$ must be perpendicular to the plane. Therefore, the position vector $$\vec{OP}$$ serves as the normal vector ($$\vec{n}$$) to the plane.

$$\vec{OP} = (4-0)\hat{i} + (3-0)\hat{j} + (5-0)\hat{k} = 4\hat{i} + 3\hat{j} + 5\hat{k}$$

The direction ratios of the normal are $$(4, 3, 5)$$

Equation of plane: $$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$

$$4(x - 4) + 3(y - 3) + 5(z - 5) = 0$$

$$4x + 3y + 5z = 50$$