The length of the perpendicular from the point $$(1, -2, 5)$$ on the line passing through $$(1, 2, 4)$$ and parallel to the line $$x + y - z = 0 = x - 2y + 3z - 5$$ is:

We need the direction of the line of intersection of the planes $$x + y - z = 0$$ and $$x - 2y + 3z - 5 = 0$$. This direction is given by the cross product of their normal vectors.

Normal to the first plane: $$\vec{n_1} = (1, 1, -1)$$. Normal to the second plane: $$\vec{n_2} = (1, -2, 3)$$.

$$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -2 & 3 \end{vmatrix}$$

$$= \hat{i}(1 \cdot 3 - (-1)(-2)) - \hat{j}(1 \cdot 3 - (-1) \cdot 1) + \hat{k}(1 \cdot (-2) - 1 \cdot 1)$$

$$= \hat{i}(3 - 2) - \hat{j}(3 + 1) + \hat{k}(-2 - 1) = (1, -4, -3)$$

The line passes through $$A = (1, 2, 4)$$ with direction vector $$\vec{d} = (1, -4, -3)$$. We need the perpendicular distance from $$P = (1, -2, 5)$$ to this line.

$$\vec{AP} = P - A = (1 - 1, -2 - 2, 5 - 4) = (0, -4, 1)$$

The perpendicular distance formula is $$\text{distance} = \dfrac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$$.

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 1 \\ 1 & -4 & -3 \end{vmatrix}$$

$$= \hat{i}((-4)(-3) - (1)(-4)) - \hat{j}((0)(-3) - (1)(1)) + \hat{k}((0)(-4) - (-4)(1))$$

$$= \hat{i}(12 + 4) - \hat{j}(0 - 1) + \hat{k}(0 + 4) = (16, 1, 4)$$

$$|\vec{AP} \times \vec{d}|^2 = 16^2 + 1^2 + 4^2 = 256 + 1 + 16 = 273$$

$$|\vec{d}|^2 = 1^2 + (-4)^2 + (-3)^2 = 1 + 16 + 9 = 26$$

$$\text{distance}^2 = \dfrac{273}{26} = \dfrac{21}{2}$$

$$\text{distance} = \sqrt{\dfrac{21}{2}}$$

The answer is Option A: $$\sqrt{\dfrac{21}{2}}$$.