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

$6$

The correct answer is Option (3) → $6$ ##

We have, $\cos^{-1} \alpha + \cos^{-1} \beta + \cos^{-1} \gamma = 3\pi$

We know that, $0 \le \cos^{-1} x \le \pi$

$⇒\cos^{-1} \alpha + \cos^{-1} \beta + \cos^{-1} \gamma = 3\pi$ is possible if and only if,

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

$⇒\cos \pi = \alpha = \beta = \gamma$

$⇒\alpha = \beta = \gamma = -1$

$∴\alpha(\beta + \gamma) + \beta(\gamma + \alpha) + \gamma(\alpha + \beta) = -1(-1-1) - 1(-1-1) - 1(-1-1)$

$= 2 + 2 + 2 = 6$