(c)
We have $\mathcal{z} = 0$ for the point, where the line intersects the curve
Therefore, $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{0 - 1}{- 1}$
$\Rightarrow \frac{x - 2}{3} = 1$ and $\frac{y + 1}{2} = 1$
$\Rightarrow x = 5$ and $y = 1$
Putting these values in $xy = c^{2}$, we get
$5 = c^{2} \Rightarrow \ c = \pm \sqrt{5}$