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Write as a system of 3 equations: 

$3a = b+c$  (Equation 1)

$5b = a+c$  (Equation 2)

$c=80+b$  (Equation 3)

Sub in (Equation 3) into (Equation 2):  $5b = a+b+80$

Simplify:  $b = {a \over 4}+20$

Now, sub what we found for b, and (Equation 3) into (Equation 1): $3a= {a \over 4} +20+80+ {a \over 4} +20$

Solving, we find $a = 48$

Subbing the value of $a$ we just found and (Equation 3) into (Equation 2), we get: $5b = 48+80+b$

Solving, we find $b = 32$

Plugging in what we found for the value of $b$ into (Equation 3), we find that $c = 112$.

Now do $a + b+c$ to find your answer. 

From BB:

a = (b+c)/3     and   a = 5b - c       <======equate

 5b-c = (b+c)/3

   re-arrange to    16b - 4c = 0         sub in  c = 80 + b

                             16 b - 4(80+b) = 0            results in    b = 32      then   c = 112      then a = 48      summed = 192