The line $$L_1$$ is parallel to the vector $$\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$$ and passes through the point $$(7, 6, 2)$$ and the line $$L_2$$ is parallel to the vector $$\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$$ and passes through the point $$(5, 3, 4)$$. The shortest distance between the lines $$L_1$$ and $$L_2$$ is :

Let $$L_1$$ pass through $$A(7,\,6,\,2)$$ with direction vector $$\vec{a}= -3\hat{i}+2\hat{j}+4\hat{k}$$ and $$L_2$$ pass through $$B(5,\,3,\,4)$$ with direction vector $$\vec{b}= 2\hat{i}+\hat{j}+3\hat{k}$$.

The shortest distance $$d$$ between two skew lines is given by
$$d=\frac{\bigl|(\vec{AB}\,.\,(\vec{a}\times\vec{b}))\bigr|}{\lVert\vec{a}\times\vec{b}\rVert}$$
where $$\vec{AB}=\overrightarrow{BA}$$ or $$\overrightarrow{AB}$$ is the vector joining any point on one line to any point on the other.

Step 1 : Vector joining the two reference points
$$\vec{AB}= \bigl(5-7\bigr)\hat{i}+\bigl(3-6\bigr)\hat{j}+\bigl(4-2\bigr)\hat{k}= -2\hat{i}-3\hat{j}+2\hat{k}$$ $$-(1)$$

Step 2 : Cross product of the direction vectors
$$\vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ -3 & 2 & 4\\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(2\cdot3-4\cdot1)\;-\;\hat{j}((-3)\cdot3-4\cdot2)\;+\;\hat{k}((-3)\cdot1-2\cdot2)$$ $$= 2\hat{i}+17\hat{j}-7\hat{k}$$ $$-(2)$$

Step 3 : Magnitude of the cross product
$$\lVert\vec{a}\times\vec{b}\rVert= \sqrt{2^{2}+17^{2}+(-7)^{2}} =\sqrt{4+289+49} =\sqrt{342} =3\sqrt{38}$$ $$-(3)$$

Step 4 : Scalar triple product
$$(\vec{AB}\,.\,(\vec{a}\times\vec{b}))= (-2)(2)+(-3)(17)+(2)(-7) =-4-51-14 =-69$$
Hence $$\bigl|(\vec{AB}\,.\,(\vec{a}\times\vec{b}))\bigr|=69$$ $$-(4)$$

Step 5 : Shortest distance
Substituting $$(3)$$ and $$(4)$$ into the formula, we get
$$d=\frac{69}{3\sqrt{38}} =\frac{23}{\sqrt{38}}$$ $$-(5)$$

Thus the shortest distance between $$L_1$$ and $$L_2$$ is $$\dfrac{23}{\sqrt{38}}$$.

Option A is correct.