To solve the quadratic equation \((1 - 3i)z^2 - (2 + i)z + i = 0\), we use the quadratic formula:

\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1 - 3i\), \(b = -(2 + i)\), and \(c = i\).

First, calculate \(b^2 - 4ac\):

\(b^2 = (-(2 + i))^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i\)

\(4ac = 4(1 - 3i)(i) = 4(i - 3i^2) = 4(i + 3) = 12 + 4i\)

\(b^2 - 4ac = (3 + 4i) - (12 + 4i) = -9\)

Now, substitute into the quadratic formula:

\(z = \frac{2 + i \pm \sqrt{-9}}{2(1 - 3i)}\)

\(\sqrt{-9} = 3i\), so:

\(z = \frac{2 + i \pm 3i}{2(1 - 3i)}\)

This gives two solutions:

\(z = \frac{2 + 4i}{2(1 - 3i)} \quad \text{and} \quad z = \frac{2 - 2i}{2(1 - 3i)}\)

Multiply numerator and denominator by the conjugate \((1 + 3i)\):

For \(z = \frac{2 + 4i}{2(1 - 3i)}\):

\(z = \frac{(2 + 4i)(1 + 3i)}{2((1 - 3i)(1 + 3i))} = \frac{2 + 6i + 4i + 12i^2}{2(1 + 9)} = \frac{2 + 10i - 12}{20} = \frac{-10 + 10i}{20} = -\frac{1}{2} + \frac{1}{2}i\)

For \(z = \frac{2 - 2i}{2(1 - 3i)}\):

\(z = \frac{(2 - 2i)(1 + 3i)}{2((1 - 3i)(1 + 3i))} = \frac{2 + 6i - 2i - 6i^2}{20} = \frac{2 + 4i + 6}{20} = \frac{8 + 4i}{20} = \frac{2}{5} + \frac{1}{5}i\)