If $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube, where $$p, q, r$$ and $$s$$ are positive integers, then the smallest value of $$p + q + r + s$$ is :

It has been given that $$N = (11^{p + 7})(7^{q - 2})(5^{r + 1})(3^{s})$$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0. 

$$p,q,r$$ and $$s$$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $$0$$.

$$11^{p + 7}$$:

This expression must be made a perfect cube. The nearest perfect cube is $$11^9$$. Therefore, the least value that $$p$$ can take is $$9-7=2$$.

$$7^{q - 2}$$

The least value that $$q$$ can take is 2. If $$q=2$$, then the value of the expression $$7^{q-2}$$ will become $$7^0=1$$, without preventing the product from becoming a perfect cube. 

$$5^{r+1}$$:

The least value that $$r$$ can take is $$2$$.

$$3^{s})$$:

The least value that $$s$$ can take is $$3$$. 

Therefore, the least value of the expression $$p + q + r + s$$ is $$2+2+2+3=9$$.
Therefore, option E is the right answer.