The line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}$ intersects the curve $xy = c^2 , z= 0 $, if c = 

$±\sqrt{5}$

At the point on the line where it intersects the given curve, we have z = 0, so that

$\frac{x-2}{3}=\frac{y+1}{2}=\frac{0-1}{-1}$

$⇒ \frac{x-2}{3}=1$ and $ \frac{y+1}{2}=1⇒ x = 5 $ and $y =1.$

Putting these values of x and y in $xy = c^2 $, we get $c^2 = 5 ⇒ c = ±\sqrt{5}$