Obviously AM GM

A similar approach would be to use the Cauchy Schwarz Inequality: $(a^{2} + a^{2} +4 + a^{2}-4)(\tan^{2}A+\tan^{2}B + \tan^{2}C) \ge (6a)^{2}$ $min(\tan^{2}A+\tan^{2}B + \tan^{2}C)= 12$

$For\quad such\quad type\quad of\quad question\quad it\quad is\quad better\\ to\quad use\quad vector\quad inequalities.\\ Assume\quad two\quad vectors\quad such\quad that\\ \\ \overrightarrow { p } \quad =\quad \tan { A\hat { \imath } } \quad +\quad \tan { B } \hat { \jmath } \quad +\quad \tan { C } \hat { k } \\ \overrightarrow { q } \quad =\quad a\hat { \imath } \quad +\quad \sqrt { { { a }^{ 2 } }{ \quad +\quad 4 } } \hat { \jmath } \quad +\quad \sqrt { { { a }^{ 2 } }{ \quad -\quad 4 } } \hat { k } \\ \\ Now\quad using\quad inequality\quad -\left| \overrightarrow { p } \right| .\left| \overrightarrow { q } \right| \quad \le \quad \overrightarrow { p } .\overrightarrow { q } \\ \\ -\sqrt { \tan ^{ 2 }{ A } +\tan ^{ 2 }{ B } +\tan ^{ 2 }{ C } } .\sqrt { { a }^{ 2 }{ +a }^{ 2 }{ +4+a }^{ 2 }-4 } \quad \le \quad 6a\\ \\ Solving\quad this\quad we\quad get\quad minimum\quad value\quad =\quad 12$