jmc

Since $p,q,r$ are roots of $30x^3-50x^2+22x-1$, $1-p,1-q,1-r$ are roots of $30(1-x{)}^3-50(1-x{)}^2+22(1-x)-1$.

If we consider only the constant terms in the expansion of the above polynomial, we find that the constant coefficient is $30-50+22-1=1$. Similarly, the linear coefficient of the above polynomial is $30(-3)+50(2)-22=-12$

Hence, $\frac{1}{1-p},\frac{1}{1-q},\frac{1}{1-r}$ are the roots of a cubic in the reversed form $1x^3-12x^2+\cdots$. Using Vieta's formula, $$\frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=-\frac{-12}{1}=\boxed{12}.$$