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

Equations of the Lines:

Equation of $ L_1 $:

$ \mathbf{r} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k} + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k}) $

Equation of $ L_2 $:

$ \mathbf{r} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} + \mu(2 \hat{i} + \hat{j} + 3 \hat{k}) $

Shortest Distance Between $ L_1 $ and $ L_2 $:

The formula for the shortest distance between two skew lines is given by:

$ \text{Distance} = \left|\frac{\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| $

Calculations:

Vector between the points:

$ \vec{a}_2 - \vec{a}_1 = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) = -2 \hat{i} - 3 \hat{j} + 2 \hat{k} $

Cross product of direction vectors:

$ \vec{b}_1 = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}, \quad \vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k} $

$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} = 2 \hat{i} + 17 \hat{j} - 7 \hat{k} $

Magnitude of the cross product:

$ \left|\vec{b}_1 \times \vec{b}_2\right| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{342} $

Dot product:

$ \left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right) = (-2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k}) $

$ = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69 $

Shortest Distance:

$ \text{Distance} = \left|\frac{-69}{\sqrt{342}}\right| = \frac{23}{\sqrt{38}} $