Solution:

We can multiply each side by $$(s - p)(s - q)(s - r)$$ to get $$A(s - q)(s - r) + B(s - p)(s - r)$$$$+ C(s - p)(s - q) = 1.$$

We can expand to get $$s^2(A + B + C) - s \cdot$$ $$(Aq + Ar + Bp + Br + Cp + Cq)$$ $$+ (Aqr + Bpr + Cpq - 1) = 0.$$

Note that the coefficients of $s$ and $s^2$ must both be $0.$

From $$A + B + C = 0, \tag*{(1)}$$ we get $$A = -(B + C),$$$$B = -(A + C),$$ and $$C = -(A + B).$$

Plugging this into the coefficient of $s,$ we get $$Ap + Bq + Cr = 0.$$

Subtracting $(1) \cdot r$ from this equation, we get $$A(p - r) + B(q - r) = 0 (2)$$ We also know that $$Aqr + Bpr + Cpq = 1.$$ Subtracting $(1) \cdot pq$ from this equation, we get $$Aq(r - p) + Bp(r - q) = 1.$$

Adding $(2) \cdot p$ to this equation, we get $$A(r - p)(q - p) = 1,$$ which gets us $$A = \dfrac{1}{(r - p)(q - p)}.$$

Similarly, we get $$B = \dfrac{1}{(r - q)(p - q)}$$ and $$C = \dfrac{1}{(q - r)(p - r)}.$$

This gives us $$\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} =$$ $$p^2 + q^2 + r^2 - pq - qr - pr.$$

Using Vieta's formulas, we get $$p^2 + q^2 + r^2 = (p + q + r)^2 -$$ $$2(pq + qr + pr) = 324.$$ Finally, since $$pq + qr + pr = 80,$$ we get our desired value of $324 - 80 = 244.$

Thus, B is the correct answer.