Calculate the determinant of the matrix.

Let the given matrix be $A$:

The determinant of a $3 \times 3$ matrix $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is given by $a(ei - fh) - b(di - fg) + c(dh - eg)$.

Applying this formula to our matrix:

1. The first minor:

   $$\left| \begin{matrix} 3i & -1 \\ 3 & i \end{matrix} \right| = (3i)(i) - (-1)(3) = 3i^2 + 3$$

   Since $i^2 = -1$:

   $$3(-1) + 3 = -3 + 3 = 0$$

2. The second minor:

   $$\left| \begin{matrix} 4 & -1 \\ 20 & i \end{matrix} \right| = (4)(i) - (-1)(20) = 4i + 20$$

3. The third minor:

   $$\left| \begin{matrix} 4 & 3i \\ 20 & 3 \end{matrix} \right| = (4)(3) - (3i)(20) = 12 - 60i$$

Distribute $3i$:

Substitute $i^2 = -1$:

Combine the real and imaginary parts:

From $x + iy = 0$, we can equate the real and imaginary parts:
$x = 0$
$y = 0$

Now, we need to find $x - iy$:

Comparing this result with the given options:
a. 0
b. $1 + 3i$
c. $3i$
d. $3 + i$

The calculated value matches option a.

The final answer is $\boxed{\text{a. 0}}$.