If $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = 3$$, then $$\frac{2a^2 + 3c^2 + 4e^2}{2b^2 + 3d^2 +4f^2}$$ = ?
Putting a=3b , c=3d , e=3f in given equation $$\frac{2a^2 + 3c^2 + 4e^2}{2b^2 + 3d^2 +4f^2}$$
It will get reduce to $$\frac{18b^2 + 27d^2 + 36f^2}{2b^2 + 3d^2 +4f^2}$$
or $$\frac{9(2b^2 + 3d^2 + 4f^2)}{2b^2 + 3d^2 +4f^2}$$ = 9
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