Number of integral terms in the expansion of $$\left\{7^{(1/2)} + 11^{(1/6)}\right\}^{824}$$ is equal to ______.

Consider the expansion of $$(7^{1/2} + 11^{1/6})^{824}$$. The general term of this expansion is given by $$ T_{r+1} = \binom{824}{r} (7^{1/2})^{824-r} (11^{1/6})^r = \binom{824}{r} 7^{(824-r)/2} \cdot 11^{r/6} $$.

For such a term to be integral, both exponents must be integers. The condition $$\frac{824 - r}{2} \in \mathbb{Z}$$ implies that $$r$$ must be even, while $$\frac{r}{6} \in \mathbb{Z}$$ implies that $$r$$ must be divisible by 6.

Hence $$r$$ must be a multiple of 6 subject to $$0 \leq r \leq 824$$, namely $$r = 0, 6, 12, \ldots, 822$$.

The total number of integral terms is given by $$\frac{822 - 0}{6} + 1 = 137 + 1 = 138$$.

Therefore, the answer is $$\boxed{138}$$.