Solution

The equations of the given lines are
$$\vec { r } =\left( \hat { i }

+2\hat { j } +\hat { k } \right) +\lambda \left( \hat { i } -\hat { j }

+\hat { k } \right) $$ and
$$\vec { r } =\hat { 2i } -\hat { j } -\hat { k } +\mu \left( \hat { 2i } +\hat { j } +2\hat { k } \right) $$
It is known that the shortest distance between the lines $$\vec { r }

=\vec { { a }_{ 1 } } +\lambda \vec { { b }_{ 1 } } $$ and $$\vec { r }

=\vec { { a }_{ 2 } } +\mu \vec { { b }_{ 2 } } $$ is given by,
$$d=\left|

\cfrac { \left( \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } }

\right) \left( \vec { { a }_{ 1 } } -\vec { { a }_{ 2 } } \right) }{

\left| \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } \right| }

\right| .....(1)$$
Comparing the given equations, we obtain
$$\vec { { a }_{ 1 } } =\hat { i } +2\hat { j } +\hat { k } $$
$$\vec { { b }_{ 1 } } =\hat { i } -\hat { j } +\hat { k } $$
$$\vec { { a }_{ 2 } } =2\hat { i } -\hat { j } -\hat { k } $$
$$\vec { { b }_{ 2 } } =2\hat { i } +\hat { j } +2\hat { k } $$
$$\vec

{ { a }_{ 2 } } -\vec { { a }_{ 1 } } =\left( 2\hat { i } -\hat { j }

-\hat { k } \right) -\left( \hat { i } +2\hat { j } +\hat { k }

\right) =\left( \hat { i } -3\hat { j } -2\hat { k } \right) $$
$$\vec

{ { b }_{ 1 } } \times \vec { { b }_{ 2 } } =\begin{vmatrix} \hat { i }

& \hat { j } & \hat { j } \\ 1 & -1 & 1 \\ 2 & 1

& 2 \end{vmatrix}$$
$$\vec { { b }_{ 1 } } \times \vec { { b }_{

2 } } =(-2-1)\hat { i } -(2-2)\hat { j } +(1+2)\hat { k } =-3\hat { i }

+3\hat { k } $$
$$\Rightarrow$$ $$\left| \vec { { b }_{ 1 } } \times

\vec { { b }_{ 2 } } \right| =\sqrt { { \left( -3 \right) }^{ 2 }+{

\left( 3 \right) }^{ 2 } } =\sqrt { 18 } =3\sqrt { 2 } $$
Substituting all the values in equation (1), we obtain
$$d=\left|

\cfrac { \left( \hat { 3\hat { i } } +3\hat { k } \right) .\left(

\hat { i } -3\hat { j } -2\hat { k } \right) }{ 3\sqrt { 2 } }

\right| $$
$$\Rightarrow$$ $$d=\left| \cfrac { -3.1+3(-2) }{ 3\sqrt { 2 } } \right| $$
$$\Rightarrow$$ $$d=\left| \cfrac { -9 }{ 3\sqrt { 2 } } \right| $$
$$\Rightarrow$$ $$d=\cfrac{3}{\sqrt 2}=\cfrac{3\times \sqrt {2}}{\sqrt {2} \times \sqrt {2}}=\cfrac {3\sqrt 2}{2}$$
Therefore, the shortest distance between the two lines is $$\cfrac{3\sqrt 2}{2}$$ units.