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Expression : $$\ \sqrt{7x+12}+\sqrt{7x-12}=3+\sqrt{33}\ $$
$$(7x-12)$$ must be greater than equal to 0.
=> $$7x-12\geq0$$
=> $$x\geq\frac{12}{7}$$
Thus, first two options are eliminated. Putting $$x=3$$ in above equation,
=> $$\sqrt{7(3)+12}+\sqrt{7(3)-12}$$
=> $$\sqrt{33}+\sqrt9$$
= $$3+\sqrt{33}=$$ R.H.S.
=> Ans - (C)
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