A square is inscribed in the circle $$x^2 + y^2 - 10x - 6y + 30 = 0$$. One side of this square is parallel to y = x + 3. If $$(x_i, y_i)$$ are the vertices of the square, then $$\sum(x_i^2 + y_i^2)$$ is equal to:

The given circle is $$x^2 + y^2 - 10x - 6y + 30 = 0$$.
To locate its centre and radius we complete the squares.

$$(x^2 - 10x) + (y^2 - 6y) + 30 = 0$$
Add and subtract $$25$$ in the first bracket and $$9$$ in the second bracket:

$$(x^2 - 10x + 25) + (y^2 - 6y + 9) + 30 - 25 - 9 = 0$$

$$(x - 5)^2 + (y - 3)^2 - 4 = 0$$

Hence the centre is $$C(5,\,3)$$ and the radius is $$r = 2$$.

For a square inscribed in a circle, the diagonals are diameters of the circle. Therefore the point of intersection of the diagonals is the centre of the circle, i.e. $$C(5,3)$$ is also the centre of the square.

If the side length of the square is $$s$$ and its circum-radius is $$R$$, the relation is $$R = \dfrac{\text{diagonal}}{2} = \dfrac{s\sqrt{2}}{2} = \dfrac{s}{\sqrt{2}}$$ $$-(1)$$

Putting $$R = r = 2$$ in $$(1)$$: $$s = R\sqrt{2} = 2\sqrt{2}$$.

One side of the square is parallel to $$y = x + 3$$, whose slope is $$1$$. A square whose sides have slopes $$\pm 1$$ has its diagonals along the coordinate axes. Thus, starting from the centre $$C(5,3)$$, the vertices are reached by moving a distance $$R = 2$$ along the positive and negative coordinate axes.

Therefore the four vertices are $$P_1(5+2,\,3) = (7,\,3),\; P_2(5-2,\,3) = (3,\,3),$$ $$P_3(5,\,3+2) = (5,\,5),\; P_4(5,\,3-2) = (5,\,1).$$

Now compute $$x_i^2 + y_i^2$$ for each vertex:

For $$P_1(7,3):\; 7^2 + 3^2 = 49 + 9 = 58$$

For $$P_2(3,3):\; 3^2 + 3^2 = 9 + 9 = 18$$

For $$P_3(5,5):\; 5^2 + 5^2 = 25 + 25 = 50$$

For $$P_4(5,1):\; 5^2 + 1^2 = 25 + 1 = 26$$

The required sum is $$\sum (x_i^2 + y_i^2) = 58 + 18 + 50 + 26 = 152.$$

Hence the correct option is Option B: $$152$$.