\(\dfrac ab=\dfrac38\implies a=\dfrac38b\)
\(\dfrac bc=\dfrac6{11}\implies c=\dfrac{11}6b\)
\(\implies a+b+c=\left(\dfrac38+1+\dfrac{11}6\right)b=\dfrac{77}{24}b\)
Since each of \(a,b,c\) are integers, so must be \(a+b+c=\dfrac{77}{24}b\). The only way for this to be the case is if \(b\) is a multiple of 24 because 77 and 24 share no common divisors. The smallest multiple would be \(b=24\). So the smallest value of \(a+b+c\) is 77.
