If ' $a$ ' is real constant and A, B, C are variable angles and, $\sqrt{a^2-4} \tan A+a \tan B+\sqrt{a^2+4} \tan C=6 a$ then the least value of $\tan ^2 A+\tan ^2 B+\tan ^2 C$ is : 

12

The given relation can be re–written as

$\left(\sqrt{a^2-4} \hat{i}+a \hat{j}+\sqrt{a^2+4} \hat{k}\right) . (\tan A \hat{i}+\tan B \hat{j}+\tan C \hat{k})=6 a$

$\Rightarrow \sqrt{\left(a^2-4\right)+a^2+\left(a^2+4\right)} . \sqrt{\tan ^2 A+\tan ^2 B+\tan ^2 C} . \cos \theta=6 a$        (as, a.b = |a| |b| cos θ)

$\Rightarrow \sqrt{3} a . \sqrt{\tan ^2 A+\tan ^2 B+\tan ^2 C} . \cos \theta=6 a$

$\Rightarrow \tan ^2 A+\tan ^2 B+\tan ^2 C=12 \sec ^2 \theta$      .......(i)

also, $12 \sec ^2 \theta \geq 12$     (as, $\sec ^2 \theta \geq 1$)    .......(ii)

from (i) and (ii), $\tan ^2 A+\tan ^2 B+\tan ^2 C \geq 12$

Hence least value of $\tan ^2 A+\tan ^2 B+\tan ^2 C=12$

Hence (3) is correct answer.