To find the shortest distance between two skew lines, we use the formula:
\[
d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot \vec{n}|}{|\vec{n}|}
\]
Where:
- \(\vec{a_1}\) and \(\vec{a_2}\) are points on the lines.
- \(\vec{b_1}\) and \(\vec{b_2}\) are direction vectors of the lines.
- \(\vec{n} = \vec{b_1} \times \vec{b_2}\) is the normal vector.
First, identify the given equations of the lines:
\[
\vec{r_1} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})
\]
\[
\vec{r_2} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})
\]
Then, extract:
- \(\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}\)
- \(\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}\)
- \(\vec{b_1} = \hat{i} - 3\hat{j} + 2\hat{k}\)
- \(\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}\)
Calculate the cross product \(\vec{b_1} \times \vec{b_2}\):
\[
\vec{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 2 \\
2 & 3 & 1
\end{vmatrix}
= \hat{i}( -3 \times 1 - 2 \times 3) - \hat{j}( 1 \times 1 - 2 \times 2 ) + \hat{k}( 1 \times 3 - (-3) \times 2 )
\]
\[
\vec{n} = -9\hat{i} + 3\hat{j} + 9\hat{k}
\]
Simplify it to \(\vec{n} = -9\hat{i} + 3\hat{j} + 9\hat{k}\).
Now calculate \(\vec{a_2} - \vec{a_1}\):
\[
\vec{a_2} - \vec{a_1} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}
\]
Find the dot product \((\vec{a_2} - \vec{a_1}) \cdot \vec{n}\):
\[
(3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = 3(-9) + 3(3) + 3(9) = -27 + 9 + 27 = 9
\]
Calculate the magnitude of \(\vec{n}\):
\[
|\vec{n}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171}
\]
Thus, the shortest distance \(d\) is:
\[
d = \frac{|9|}{\sqrt{171}} = \frac{9}{\sqrt{171}} = \frac{9}{\sqrt{9 \cdot 19}} = \frac{3}{\sqrt{19}}
\]
Hence, the shortest distance between the two skew lines is \(\frac{3}{\sqrt{19}}\).