>>829
To solve this problem, we need to determine the elements of the set \( S \) that are also in the set \( Q \), which contains all rational numbers. Let's analyze each element in \( S \):

1. \(\left( \frac{-1 + i \sqrt{3}}{2} \right)^{2019}\):
   This is a complex number raised to a high power. Notice that \(\frac{-1 + i \sqrt{3}}{2}\) is a complex number on the unit circle in the complex plane. Specifically, it represents the cube root of unity, \( e^{2\pi i / 3} \). Raising this to the 2019th power will give another root of unity, but not a rational number.

2. \(\Im \left( \left( \frac{1 + i}{\sqrt{2}} \right)^{2019} \right)\):
   The imaginary part of a complex number raised to a high power. Since \(\frac{1 + i}{\sqrt{2}}\) is on the unit circle, raising it to the 2019th power will rotate it on the unit circle, but its imaginary part will generally not be rational.

3. \(\frac{1}{3}\):
   This is a rational number.

4. \(\frac{\pi}{3}\):
   This is not a rational number because \(\pi\) is irrational.

5. \(\sin \frac{\pi}{3}\):
   \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\), which is not rational.

6. \(\frac{22}{7}\):
   While \(\frac{22}{7}\) is a common approximation of \(\pi\), it is a rational number.

Now, let's count the rational numbers in \( S \):
- \(\frac{1}{3}\)
- \(\frac{22}{7}\)

There are 2 rational numbers in \( S \).

Therefore, \(|S \cap Q|\) equals 2.

The answer is \(2BM3FRT\).