If 3a=4b=6c and a+b+c=2729 , then a2+b2+c2 =?

$4b=6c\Rightarrow b=\frac{3}{2}c and 3a=4b\Rightarrow a=\frac{4}{3}b=\frac{4}{3}\left(\frac{3}{2}c\right)=2c$ 

$a+b+c=2729 \Rightarrow 2c+\frac{3}{2}c+c=2729 \Rightarrow c=629$

  $\therefore a2+b2+c2=(a+b+b)2-2(ab+bc+ca)$ 

$=\left(2729\right)2-22c\times \frac{3}{2}c+\frac{3}{2}c\times c+c\times 2c$ 

  $=729\times 29-2\times \frac{13}{2} c2$

= 87