Let $$P(a, b)$$ be a point on the parabola $$y^2 = 8x$$ such that the tangent at $$P$$ passes through the centre of the circle $$x^2 + y^2 - 10x - 14y + 65 = 0$$. Let $$A$$ be the product of all possible values of $$a$$ and $$B$$ be the product of all possible values of $$b$$. Then the value of $$A + B$$ is equal to

Given parabola,

$$y^2=8x$$

Comparing with

$$y^2=4ax,$$

we get

$$a=2$$

Let

$$P(a,b)$$

lie on the parabola.

Then,

$$b^2=8a\qquad\cdots(1)$$

Equation of tangent at point $$(a,b)$$ to

$$y^2=8x$$

is

$$by=4(x+a)$$

$$4x-by+4a=0$$

Now find the centre of the circle

$$x^2+y^2-10x-14y+65=0$$

Comparing with

$$x^2+y^2+2gx+2fy+c=0,$$

we get

$$g=-5,\qquad f=-7$$

Hence centre is

$$(5,7)$$

Since the tangent passes through the centre,

$$4(5)-b(7)+4a=0$$

$$20-7b+4a=0$$

$$4a-7b+20=0\qquad\cdots(2)$$

Using (1),

$$a=\frac{b^2}{8}$$

Substitute in (2):

$$4\left(\frac{b^2}{8}\right)-7b+20=0$$

$$\frac{b^2}{2}-7b+20=0$$

$$b^2-14b+40=0$$

$$(b-4)(b-10)=0$$

Hence,

$$b=4\quad \text{or}\quad b=10$$

Corresponding $$a$$ values are

$$a=\frac{4^2}{8}=2$$

and

$$a=\frac{10^2}{8}=\frac{25}{2}$$

Therefore,

$$A=2\cdot\frac{25}{2}=25$$

and

$$B=4\cdot10=40$$

Hence,

$$A+B=25+40=65$$

Therefore, the required value is

$$\boxed{65}$$.