Let the direction cosines of line $$L$$ be $$l,m,n,$$

then $$2l+3m+n=0$$ $$\,\,\,\,\,\,\,....\left( i \right)$$

and $$l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)$$

on solving equation $$(i)$$ and $$(ii),$$ we get

$${l \over {6 - 3}} = {m \over {1 - 4}} = {n \over {6 - 3}}$$

$$\,\,\,\,\,\,\,\,\,\, \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3}$$

Now $$ \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3} = {{\sqrt {{l^2} + {m^2} + {n^2}} } \over {\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }}$$

As $${l^2} + {m^2} + {n^2} = 1$$

$$\therefore$$ $${l \over 3} = {m \over { - 3}} = {n \over 3} = {1 \over {\sqrt {27} }}$$

$$ \Rightarrow l = {3 \over {\sqrt {27} }} = {1 \over {\sqrt 3 }},\,\,m = - {1 \over {\sqrt 3 }},n = {1 \over {\sqrt 3 }}$$

Line $$L,$$ makes an angle $$\alpha $$ with $$+ve$$ $$x$$-axis

$$\therefore$$ $$l = \cos \,\alpha \,\,\,\, \Rightarrow \,\,\,\cos \alpha \,\, = {1 \over {\sqrt 3 }}$$