Find the magnitude of the shortest distance between the lines $$\frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{1} \text {and} \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2} .$$​

Correct option is A

​​​​​$$\begin{aligned}&\text{From given data:} \\&\quad \text{A point on Line 1: } \vec{A} = (0, 0, 0), \quad \text{Direction vector of Line 1: } \vec{d_1} = (2, -3, 1) \\&\quad \text{A point on Line 2: } \vec{B} = (2, 1, -2), \quad \text{Direction vector of Line 2: } \vec{d_2} = (3, -5, 2) \\&\text{Shortest distance between the two skew lines:} \\&D = \frac{\left| (\vec{B} - \vec{A}) \cdot (\vec{d_1} \times \vec{d_2}) \right|}{|\vec{d_1} \times \vec{d_2}|} \\&\text{Compute } \vec{B} - \vec{A} = (2, 1, -2) - (0, 0, 0) = (2, 1, -2) \\&\vec{d_1} \times \vec{d_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\2 & -3 & 1 \\3 & -5 & 2\end{vmatrix} \\&= \hat{i} \left( (-3)(2) - (1)(-5) \right) - \hat{j} \left( (2)(2) - (1)(3) \right) + \hat{k} \left( (2)(-5) - (-3)(3) \right) \\&= \hat{i}(-1) - \hat{j}(1) + \hat{k}(-1) = (-1, -1, -1) \ \\&(\vec{B} - \vec{A}) \cdot (\vec{d_1} \times \vec{d_2}) = (2, 1, -2) \cdot (-1, -1, -1) = -2 - 1 + 2 = -1 \Rightarrow | -1 | = 1 \\&\text{Magnitude of cross product:} |\vec{d_1} \times \vec{d_2}| = \sqrt{3} \\&\text{The shortest distance:} \\&D = \frac{1}{\sqrt{3}} \quad \text{units}\end{aligned}$$​​