jmc

Consider the expression $P(a)=(a+b+c{)}^3-a^3-b^3-c^3$ as a polynomial in $a$. It follows that $P(-b)=(b-b+c{)}^3-(-b{)}^3-b^3-c^3=0$, so $a+b$ is a factor of the polynomial $P(a)$. By symmetry, $(a+b)(b+c)(c+a)$ divides into the expression $(a+b+c{)}^3-a^3-b^3-c^3$; as both expressions are of degree $3$ in their variables, it follows that $$(a+b+c{)}^3-a^3-b^3-c^3=k(a+b)(b+c)(c+a)=150=2\cdot 3\cdot 5\cdot 5,$$ where we can determine that $k=3$ by examining what the expansion of $(a+b+c{)}^3$ will look like. Since $a,b,$ and $c$ are positive integers, then $a+b$, $b+c$, and $c+a$ must all be greater than $1$, so it follows that $\{a+b,b+c,c+a\}=\{2,5,5\}$. Summing all three, we obtain that $$(a+b)+(b+c)+(c+a)=2(a+b+c)=2+5+5=12,$$ so $a+b+c=\boxed{6}$.