If $$P$$ and $$Q$$ are the points of intersection of the circles $$x^2 + y^2 + 3x + 7y + 2p - 5 = 0$$ and $$x^2 + y^2 + 2x + 2y - p^2 = 0$$, then there is a circle passing through $$P, Q$$ and $$(1, 1)$$ for

The two given circles are

$$S_1 : x^2 + y^2 + 3x + 7y + 2p - 5 = 0,$$
$$S_2 : x^2 + y^2 + 2x + 2y - p^2 = 0.$$

Let their points of intersection be $$P$$ and $$Q$$. (These points may be real or imaginary; the algebra that follows is valid in either case.)

Every circle that passes through the common points $$P$$ and $$Q$$ is obtained by taking a linear combination of $$S_1$$ and $$S_2$$, viz.

$$S_\lambda : \; S_1 + \lambda\,S_2 = 0,$$
where $$\lambda$$ is a real parameter.

We want a member of this family that also passes through the fixed point $$(1,1)$$. Substituting $$(1,1)$$ in the equation of $$S_\lambda$$ gives

$$S_1(1,1) + \lambda\,S_2(1,1) = 0.$$

Compute the two values separately:

$$S_1(1,1) = 1^2 + 1^2 + 3(1) + 7(1) + 2p - 5 = 7 + 2p,$$
$$S_2(1,1) = 1^2 + 1^2 + 2(1) + 2(1) - p^2 = 6 - p^2.$$

Hence the required condition is

$$\bigl(7 + 2p\bigr) + \lambda\,(6 - p^2) = 0.$$

Case 1: $$6 - p^2 \neq 0$$ (that is, $$p \neq \pm\sqrt6$$).
In this case we can choose

$$\lambda = -\,\dfrac{7 + 2p}{\,6 - p^2\,},$$

which is a perfectly finite real value. Therefore a circle through $$P,Q,(1,1)$$ exists.

Case 2: $$6 - p^2 = 0$$ (that is, $$p = \pm\sqrt6$$).
Then $$S_2(1,1)=0,$$ so the circle $$S_2 = 0$$ itself already passes through $$(1,1)$$ as well as $$P$$ and $$Q$$. Thus a suitable circle is again available (namely $$S_2$$), even though the value of $$\lambda$$ found above becomes indeterminate.

Because both cases are covered, a circle through the three points $$P,Q,(1,1)$$ exists for every real value of the parameter $$p$$.

Hence the correct option is:
Option A which is: all values of $$p$$.