Explanation

(c)
Let $\mathcal{z =}x + iy$
Then, $\mathcal{z +}i\mathcal{z =}(x + iy) + i(x + iy) = (x - y) + i(x + y)$
and $i\mathcal{z =}i(x + iy) = - y + ix$
If $\mathrm{\Delta}$ be the area of the triangle formed by $\mathcal{z,\ z +}i\mathcal{z}$ and $i\mathcal{z}$, then
$\mathrm{\Delta} = \frac{1}{2}|\left| \begin{matrix}
x & y & 1 \\
x - y & x + y & 1 \\
- y & x & 1
\end{matrix} \right||$
Applying $R_{2} \rightarrow R_{2} - (R_{1} + R_{3})$
Then $\mathrm{\Delta} = \frac{1}{2}\ \left\| \begin{matrix}
x & y & 1 \\
0 & 0 & - 1 \\
- y & x & 1
\end{matrix} \right\| = \frac{1}{2}\left( x^{2} + y^{2} \right)$
$= \frac{1}{2}\left| \mathcal{z} \right|^{2} = 200$ (given)
$\Rightarrow \ \left| \mathcal{z} \right|^{2} = 400$
$\Rightarrow \ \left| \mathcal{z} \right| = 20$
$\therefore\ \left| 3\mathcal{z} \right| = 3\left| \mathcal{z} \right| = 60$