If $$3a = 4b = 6c$$   and $$a + b + c = 27\sqrt {29} {\text{,}}$$    then $$\sqrt {{a^2} + {b^2} + {c^2}} $$    is ?

Solution (By Examveda Team)

$$\eqalign{ & a + b + c = 27\sqrt {29} \cr & \Rightarrow 2c + \frac{3}{2}c + c = 27\sqrt {29} \cr & \Rightarrow \frac{9}{2}c = 27\sqrt {29} \cr & \Rightarrow c = 6\sqrt {29} \cr & \therefore \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \cr & = \sqrt {{{\left( {27\sqrt {29} } \right)}^2} - 2\left( {2c \times \frac{3}{2}c + \frac{3}{2}c \times c + c \times 2c} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2\left( {3{c^2} + \frac{3}{2}{c^2} + 2{c^2}} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2 \times \frac{{13}}{2}{c^2}} \cr & = \sqrt {\left( {729 \times 29} \right) - 13{{\left( {6\sqrt {29} } \right)}^2}} \cr & = \sqrt {29\left( {729 - 468} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$