For $${L_1}:$$$${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$

$$\therefore$$ Point on line is $$A(2,-1,6)$$

Parallel vector to this line is,

$$\overrightarrow p = 3\widehat i + 2\widehat j + 2\widehat k$$

For $${L_2}:$$$${{x - 6} \over 3} = {{y - 1} \over -2} = {{z + 8} \over 0}$$

$$\therefore$$ Point on line is $$B(6,1,-8)$$

Parallel vector to this line is,

$$\overrightarrow q = 3\widehat i - 2\widehat j + 0\widehat k$$

Now shortest distance between this two lines is,

$$ = \left| {{{\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$$

Now, $$\overrightarrow {AB} = 4 \widehat i + 2\widehat j -14\widehat k$$

$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 2 & 2 \cr 3 & -2 & 0 \cr } } \right|$$

$$ = 4\widehat i +6\widehat j -12\widehat k$$

$$\therefore$$ $|\vec{p} \times \vec{q}|=\sqrt{16+36+144}=\sqrt{196}=14$

and $$\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right) = + 16 + 12 + 168 = +196$$

$$\therefore$$ Shortest distance $$ = \left| {{{ 196} \over {14 }}} \right| = 14$$