1. Let the arithmetic progression have first term $a$ and common difference $d=\frac{3}{2}$.

   Then the $n$-th term is $T_n=a+(n-1)d.$

2. Write the 10th and 11th terms.

   $T_{10}=a+9\cdot \frac{3}{2}=a+\frac{27}{2},$ $T_{11}=a+10\cdot \frac{3}{2}=a+15.$

   These are the roots of $3x^2-px+q=0.$

3. Use the sum of first 11 terms.

   Given $S_{11}=\frac{11}{2}\left[2a+(11-1)d\right]=88.$

   Since $d=\frac{3}{2}$, $\frac{11}{2}\left(2a+10\cdot \frac{3}{2}\right)=88.$

   $\frac{11}{2}(2a+15)=88.$

   $2a+15=16.$

   $2a=1 \implies a=\frac{1}{2}.$

4. Find the two roots explicitly.

   $T_{10}=\frac{1}{2}+\frac{27}{2}=14,$ $T_{11}=\frac{1}{2}+15=\frac{31}{2}.$

   So the roots are $14$ and $\frac{31}{2}$.

5. Use Vieta's formulas for $3x^2-px+q=0$.

   For a quadratic $ax^2+bx+c=0$:

   • sum of roots $= -\frac{b}{a}$
   • product of roots $= \frac{c}{a}$

   Here, $3x^2-px+q=0.$

   So, $\alpha+\beta=\frac{p}{3}, \qquad \alpha\beta=\frac{q}{3}.$

   Now, $\alpha+\beta=14+\frac{31}{2}=\frac{28+31}{2}=\frac{59}{2},$ hence $\frac{p}{3}=\frac{59}{2} \implies p=\frac{177}{2}.$

   Also, $\alpha\beta=14\cdot \frac{31}{2}=7\cdot 31=217,$ hence $\frac{q}{3}=217 \implies q=651.$

6. Compute $q-2p$.

   $q-2p=651-2\cdot \frac{177}{2}=651-177=474.$

Therefore, $\boxed{474}$