Let a straight line L pass through the point P(2,-1,3) and be perpendicular to the lines $$\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$$ and $$\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}.$$ If the line L intersects the yz-plane at the point Q , then the distance between the points P and Q is :

Line L passes through $$P(2,-1,3)$$ and is perpendicular to:

$$L_1$$: direction $$(2,1,-2)$$

$$L_2$$: direction $$(1,3,4)$$

Direction of L = $$(2,1,-2) \times (1,3,4)$$:

$$ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix} = \hat{i}(4+6) - \hat{j}(8+2) + \hat{k}(6-1) = (10, -10, 5) $$

Simplifying: direction $$(2, -2, 1)$$.

Line L: $$(x,y,z) = (2,-1,3) + t(2,-2,1)$$

To find intersection with yz-plane ($$x = 0$$):

$$2 + 2t = 0 \Rightarrow t = -1$$

$$Q = (0, -1+2, 3-1) = (0, 1, 2)$$

Distance PQ:

$$PQ = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4+4+1} = \sqrt{9} = 3$$

The correct answer is Option 4: 3.