Explanation

(c)
Let $\mathcal{z}_{1} = 1 + i,\ \mathcal{z}_{2} = - 2 + 3i$ and $\mathcal{z}_{3} = 0 + \frac{5}{3}i$
Then, $\left| \begin{matrix}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{matrix} \right| = \left| \begin{matrix}
1 & 1 & 1 \\
- 2 & 3 & 1 \\
0 & \frac{5}{3} & 1
\end{matrix} \right|$
$= 1\left( 3 - \frac{5}{3} \right) + 1(2) + 1\left( \frac{- 10}{3} \right)$
$= \frac{4}{3} + 2 - \frac{10}{3}$
$= \ \frac{4 + 6 - 10}{3} = 0$
Hence, area of triangle is zero, therefore points are collinear