1. General term of the binomial expansion

For $\left(7^{1/2}+11^{1/6}\right)^{824},$ the general term is $T_{r+1}=\binom{824}{r}\left(7^{1/2}\right)^{824-r}\left(11^{1/6}\right)^r,$ where $r=0,1,2,\dots,824$.

So, $T_{r+1}=\binom{824}{r}\,7^{\frac{824-r}{2}}\,11^{\frac{r}{6}}.$

2. Condition for an integral term

A term will be integral if the exponents of both $7$ and $11$ are integers.

So we need: $\frac{824-r}{2}\in \mathbb{Z}$ and $\frac{r}{6}\in \mathbb{Z}.$

3. Solve the divisibility conditions

From $\frac{r}{6}\in \mathbb{Z},$ we get $r\equiv 0 \pmod{6}.$

If $r$ is divisible by $6$, then it is automatically even, hence $824-r$ is also even because $824$ is even. Therefore, $\frac{824-r}{2}$ is automatically an integer.

So the only effective condition is: $r=6k$ for some integer $k$ with $0\le r\le 824.$

Thus, $0\le 6k\le 824 \implies 0\le k\le \left\lfloor \frac{824}{6}\right\rfloor =137.$

Hence, $k=0,1,2,\dots,137,$ which gives $137-0+1=138$ values.

4. Number of integral terms

Therefore, the number of integral terms is $\boxed{138}.$

5. Comparison with stored answer

Stored correct answer = $138$.

My derived answer also is $138$, so they agree.