jmc

We convert $n$ to base $10$. The base $7$ expression implies that $n=49A+7B+C$, and the base $11$ expression implies that $n=121C+11B+A$. Setting the two expressions equal to each other yields that $$n=49A+7B+C=121C+11B+A⟹48A-4B-120C=0.$$Isolating $B$, we get $$B=\frac{48A-120C}{4}=12A-30C=6(2A-5C).$$It follows that $B$ is divisible by $6$, and since $B$ is a base $7$ digit, then $B$ is either $0$ or $6$. If $B$ is equal to $0$, then $2A-5C=0⟹2A=5C$, so $A$ must be divisible by $5$ and hence must be either $0$ or $5$. Since $n$ is a three-digit number in base $7$, then $A\ne 0$, so $A=5$ and $C=2$. Thus, $n=502_7=5\cdot 7^2+2=247$.

If $B$ is equal to $6$, then $2A-5C=1$, so $2A-1=5C$ and $2A-1$ must be divisible by 5. Since $A$ is a base $7$ digit, it follows that $A=3$ and $C=1$. This yields the value $n=361_7=3\cdot 7^2+6\cdot 7+1=190$. The largest possible value of $n$ in base $10$ is $\boxed{247}$.