The distance of the point $$Q(0, 2, -2)$$ from the line passing through the point $$P(5, -4, 3)$$ and perpendicular to the lines $$\vec{r} = -3\hat{i} + 2\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 5\hat{k}), \lambda \in \mathbb{R}$$ and $$\vec{r} = \hat{i} - 2\hat{j} + \hat{k} + \mu(-\hat{i} + 3\hat{j} + 2\hat{k}), \mu \in \mathbb{R}$$ is

We need to find the distance of point $$Q(0,2,-2)$$ from a line through $$P(5,-4,3)$$ that is perpendicular to two given lines. Since the required line is perpendicular to both lines whose direction vectors are $$\vec{d_1} = (2, 3, 5)$$ and $$\vec{d_2} = (-1, 3, 2)$$, its direction is given by the cross product:

$$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix}$$

The component along $$\hat{i}$$ is $$(3)(2) - (5)(3) = 6 - 15 = -9,$$ along $$\hat{j}$$ is $$-((2)(2) - (5)(-1)) = -(4+5) = -9,$$ and along $$\hat{k}$$ is $$(2)(3) - (3)(-1) = 6+3 = 9.$$ Thus $$\vec{d} = (-9, -9, 9),$$ which simplifies to $$(1, 1, -1).$$

A line through $$P(5,-4,3)$$ with direction $$(1,1,-1)$$ can be parametrised as $$(x, y, z) = (5+t, -4+t, 3-t).$$ The vector from a general point on this line to $$Q$$ is $$\vec{PQ_t} = (0-(5+t), 2-(-4+t), -2-(3-t)) = (-5-t, 6-t, -5+t).$$ For the foot of the perpendicular, this vector must be perpendicular to $$\vec{d} = (1,1,-1),$$ so we set

$$(-5-t)(1) + (6-t)(1) + (-5+t)(-1) = 0$$

which simplifies to $$-5-t+6-t+5-t = 0$$ or $$6-3t = 0 \implies t = 2.$$

At $$t = 2$$, $$(5+2, -4+2, 3-2) = (7, -2, 1)$$ gives the foot of the perpendicular.

$$d = \sqrt{(0-7)^2 + (2-(-2))^2 + (-2-1)^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$$

The correct answer is Option 4: $$\sqrt{74}$$.