Let, $$\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$$

and $$\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$$

We know, Angle between two vectors

$$\cos \alpha = {{\overrightarrow {{a_1}} \,.\,\overrightarrow {{a_2}} } \over {|\overrightarrow {{a_1}} \,|.|\,\overrightarrow {{a_2}} |}}$$

$$ = {{ab + bc + ac} \over {\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt {{a^2} + {b^2} + {c^2}} }}$$

$$ = {{ab + bc + ac} \over {({a^2} + {b^2} + {c^2})}}$$

Given, $${a^2} + {b^2} + {c^2} = 1$$

$$ \therefore $$ $$\cos \alpha = ab + bc + ac$$

$$ = ab\left( {{1 \over a} + {1 \over b} + {1 \over c}} \right)$$ .....(1)

Given, $$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right) = \lambda $$ (Assume)

$$ \therefore $$ $${1 \over a} = {{\cos \theta } \over \lambda }$$

$${1 \over b} = {{\cos \left( {\theta + {{2\pi } \over 3}} \right)} \over \lambda }$$

$${1 \over c} = {{\cos \left( {\theta + {{4\pi } \over 3}} \right)} \over \lambda }$$

$$ \therefore $$ $${1 \over a} + {1 \over b} + {1 \over c} = {1 \over \lambda }\left[ {\cos \theta + \cos \left( {\theta + {{2\pi } \over 3}} \right) + \cos \left( {\theta + {{4\pi } \over 3}} \right)} \right]$$

$$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos \left( {{{2\theta + 2\pi } \over 2}} \right)\cos \left( {{{{{2\pi } \over 3}} \over 2}} \right)} \right]$$

$$ = {1 \over \lambda }\left[ {\cos \theta + 2\cos (\theta + \pi )\cos \left( {{\pi \over 3}} \right)} \right]$$

$$ = {1 \over \lambda }\left[ {\cos \theta + 2( - \cos \theta ) \times {1 \over 2}} \right]$$

$$ = {1 \over \lambda } \times 0$$

$$ = 0$$

Putting value of $$\lambda $$ in equation (1),

cos$$\alpha $$ = ab(0) = 0

$$ \Rightarrow $$ $$\alpha = {\pi \over 2}$$