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

Solution (By Examveda Team)

$$\eqalign{ & 3a = 4b = 6c \cr & \Rightarrow \frac{{3a}}{{12}} = \frac{{4b}}{{12}} = \frac{{6c}}{{12}} \Rightarrow \frac{a}{4} = \frac{b}{3} = \frac{c}{2} = k \cr & \Rightarrow a = 4k,\,b = 3k,\,c = 2k \cr & a + b + c = 27\sqrt {29} \cr & 9k = 27\sqrt {29} \cr & k = 3\sqrt {29} \cr & a = 4 \times 3\sqrt {29} ,\,b = 3 \times 3\sqrt {29} ,\,c = 2 \times 3\sqrt {29} \cr & \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {29\left( {144 + 81 + 36} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$