If each of the lines $$5x + 8y = 13$$ and $$4x - y = 3$$ contains a diameter of the circle $$x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^3 + 1 = 0$$, then :

The problem states that each of the lines $$5x + 8y = 13$$ and $$4x - y = 3$$ contains a diameter of the circle given by $$x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^3 + 1 = 0$$. For a line to contain a diameter of a circle, it must pass through the center of the circle. Therefore, the center of the circle must lie on both lines, meaning it is the point of intersection of these two lines.

To find the intersection point, solve the system of equations:

Equation 1: $$5x + 8y = 13$$

Equation 2: $$4x - y = 3$$

Solve Equation 2 for $$y$$:

$$4x - y = 3 \implies y = 4x - 3$$

Substitute $$y = 4x - 3$$ into Equation 1:

$$5x + 8(4x - 3) = 13$$

$$\implies 5x + 32x - 24 = 13$$

$$\implies 37x - 24 = 13$$

$$\implies 37x = 13 + 24 = 37$$

$$\implies x = 1$$

Then substitute $$x = 1$$ back into $$y = 4x - 3$$:

$$y = 4(1) - 3 = 4 - 3 = 1$$

So the center of the circle is $$(1, 1)$$.

The general equation of a circle is $$x^2 + y^2 + 2gx + 2fy + c = 0$$, with center $$(-g, -f)$$. Comparing this to the given circle equation $$x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^3 + 1 = 0$$, we identify:

$$2g = -2(a^2 - 7a + 11) \implies g = -(a^2 - 7a + 11) = -a^2 + 7a - 11$$

$$2f = -2(a^2 - 6a + 6) \implies f = -(a^2 - 6a + 6) = -a^2 + 6a - 6$$

The center is $$(-g, -f)$$, so:

$$-g = -(-a^2 + 7a - 11) = a^2 - 7a + 11$$

$$-f = -(-a^2 + 6a - 6) = a^2 - 6a + 6$$

Since the center is $$(1, 1)$$, we set up the equations:

$$a^2 - 7a + 11 = 1$$ ...(i)

$$a^2 - 6a + 6 = 1$$ ...(ii)

Solve equation (i):

$$a^2 - 7a + 11 = 1 \implies a^2 - 7a + 10 = 0$$

Factorizing: $$(a - 5)(a - 2) = 0 \implies a = 5$$ or $$a = 2$$

Solve equation (ii):

$$a^2 - 6a + 6 = 1 \implies a^2 - 6a + 5 = 0$$

Factorizing: $$(a - 1)(a - 5) = 0 \implies a = 1$$ or $$a = 5$$

The common solution that satisfies both equations is $$a = 5$$.

Now substitute $$a = 5$$ into the circle equation to find the condition for $$b$$. First compute the coefficients:

$$a^2 - 7a + 11 = 5^2 - 7 \times 5 + 11 = 25 - 35 + 11 = 1$$

$$a^2 - 6a + 6 = 5^2 - 6 \times 5 + 6 = 25 - 30 + 6 = 1$$

So the circle equation becomes:

$$x^2 + y^2 - 2 \times 1 \times x - 2 \times 1 \times y + b^3 + 1 = 0 \implies x^2 + y^2 - 2x - 2y + b^3 + 1 = 0$$

Rewrite this in standard form by completing the square:

$$x^2 - 2x + y^2 - 2y = -b^3 - 1$$

Add 1 and 1 to both sides to complete the squares:

$$(x^2 - 2x + 1) + (y^2 - 2y + 1) = -b^3 - 1 + 1 + 1$$

$$\implies (x - 1)^2 + (y - 1)^2 = -b^3 + 1$$

The right-hand side represents the square of the radius, which must be positive for a real circle:

$$-b^3 + 1 > 0 \implies 1 > b^3 \implies b^3 < 1$$

Since the cube function is strictly increasing, $$b^3 < 1$$ implies $$b < 1$$. Therefore, $$b \in (-\infty, 1)$$.

Now evaluate the options:

A. $$a = 5$$ and $$b \notin (-1, 1)$$ — Incorrect because $$b$$ can be in $$(-1, 1)$$ as long as $$b < 1$$ (e.g., $$b = 0$$ gives $$b^3 = 0 < 1$$).

B. $$a = 1$$ and $$b \notin (-1, 1)$$ — Incorrect because $$a = 5$$, not $$a = 1$$.

C. $$a = 2$$ and $$b \notin (-\infty, 1)$$ — Incorrect because $$a = 5$$, not $$a = 2$$, and $$b$$ must be less than 1.

D. $$a = 5$$ and $$b \in (-\infty, 1)$$ — Correct because $$a = 5$$ and $$b < 1$$.

Hence, the correct answer is Option D.