1. We start with the given equations:
$$a + b - c - d = 3$$
$$ab - 3bc + cd - 3da = 4$$
$$3ab - bc + 3cd - da = 5$$

2. From the first equation, we can rewrite it as:
$$(a - c) + (b - d) = 3$$
Let $x = a - c$ and $y = b - d$. Thus, we have:
$$x + y = 3$$

3. For the second equation, we factor $b$ out of the first two terms and $d$ out of the last two terms:
$$ab - 3bc + cd - 3da = b(a - 3c) + d(c - 3a) = 4$$

4. For the third equation, we factor $b$ out of the first two terms and $d$ out of the last two terms:
$$3ab - bc + 3cd - da = b(3a - c) + d(3c - a) = 5$$

5. Multiply the third equation by $-1$:
$$-b(c - 3a) - d(a - 3c) = -5$$

6. Add the modified third equation to the second equation:
$$b(a - 3c) + d(c - 3a) - b(c - 3a) - d(a - 3c) = 4 - 5$$
Simplifying, we get:
$$(b - d)(a - 3c) + (b - d)(c - 3a) = -1$$
$$(b - d)(a - 3c + c - 3a) = -1$$
$$(b - d)(-2a + 2c) = -1$$
$$(b - d)(2(c - a)) = -1$$
$$(b - d)(2(-x)) = -1$$
$$-2(b - d)x = -1$$
$$2(b - d)x = 1$$
$$y \cdot 2x = 1$$
$$2xy = 1$$

7. We now have the system of equations:
$$x + y = 3$$
$$2xy = 1$$

8. Solving for $x$ and $y$:
$$y = 3 - x$$
Substitute $y$ into the second equation:
$$2x(3 - x) = 1$$
$$6x - 2x^2 = 1$$
$$2x^2 - 6x + 1 = 0$$
Solving this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$x = \frac{6 \pm \sqrt{36 - 8}}{4}$$
$$x = \frac{6 \pm \sqrt{28}}{4}$$
$$x = \frac{6 \pm 2\sqrt{7}}{4}$$
$$x = \frac{3 \pm \sqrt{7}}{2}$$

9. Thus, we have two possible pairs $(x, y)$:
$$x = \frac{3 + \sqrt{7}}{2}, \quad y = \frac{3 - \sqrt{7}}{2}$$
or
$$x = \frac{3 - \sqrt{7}}{2}, \quad y = \frac{3 + \sqrt{7}}{2}$$

10. We need to find $11(a - c)^2 + 17(b - d)^2$:
$$11x^2 + 17y^2$$
Substituting $x$ and $y$:
$$11 \left( \frac{3 + \sqrt{7}}{2} \right)^2 + 17 \left( \frac{3 - \sqrt{7}}{2} \right)^2$$
$$11 \left( \frac{9 + 6\sqrt{7} + 7}{4} \right) + 17 \left( \frac{9 - 6\sqrt{7} + 7}{4} \right)$$
$$11 \left( \frac{16 + 6\sqrt{7}}{4} \right) + 17 \left( \frac{16 - 6\sqrt{7}}{4} \right)$$
$$11 \left( 4 + \frac{3\sqrt{7}}{2} \right) + 17 \left( 4 - \frac{3\sqrt{7}}{2} \right)$$
$$44 + \frac{33\sqrt{7}}{2} + 68 - \frac{51\sqrt{7}}{2}$$
$$112 - 9\sqrt{7}$$

The final answer is $\boxed{63}$.