If $$4{a^2} + 9{b^2} + 16{c^2} = 2\left( {3ab + 6bc + 4ca} \right),$$        where $$a, b, c$$  are non-zero numbers, then $$a, b, c$$  are in

Solution :
$$\eqalign{ & {\left( {2a} \right)^2} + {\left( {3b} \right)^2} + {\left( {4c} \right)^2} - \left( {2a} \right)\left( {3b} \right) - \left( {3b} \right)\left( {4c} \right) - \left( {4c} \right)\left( {2a} \right) = 0 \cr & \Rightarrow \,\,{\left( {2a - 3b} \right)^2} + {\left( {3b - 4c} \right)^2} + {\left( {4c - 2a} \right)^2} = 0 \cr & \Rightarrow \,\,2a = 3b = 4c \cr & \Rightarrow \,\,\frac{a}{{\frac{1}{2}}} = \frac{b}{{\frac{1}{3}}} = \frac{c}{{\frac{1}{4}}}. \cr} $$