If ${\cos ^{ - 1}}\alpha + {\cos ^{ - 1}}\beta + {\cos ^{ - 1}}\gamma = 3\pi$, then $\alpha (\beta + \gamma ) + \beta (\gamma + \alpha ) + \gamma (\alpha + \beta )$ equals to

We have, ${\cos ^{ - 1}}\alpha + {\cos ^{ - 1}}\beta + {\cos ^{ - 1}}\gamma = 3\pi$
We know that, $0 \le {\cos ^{ - 1}}x \le \pi$

$\Rightarrow$ ${\cos ^{ - 1}}\alpha + {\cos ^{ - 1}}\beta + {\cos ^{ - 1}}\gamma = 3\pi$

If and only if, ${\cos ^{ - 1}}\alpha = {\cos ^{ - 1}}\beta = {\cos ^{ - 1}}\gamma = \pi$

$\Rightarrow$ $\cos \pi = \alpha = \beta = \gamma$

$\Rightarrow$ $- 1 = \alpha = \beta = \gamma$

$\Rightarrow$ $\alpha = \beta = \gamma = - 1$

$therefore, \alpha (\beta + \gamma ) + \beta (\gamma + \alpha ) + \gamma (\alpha + \beta )$

$= - 1( - 1 - 1) - 1( - 1 - 1) - 1( - 1 - 1)$
$= 2 + 2 + 2 = 6$