The line $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z-1}{-1}$ intersects the curve $x^2-y^2=a^2;z=0$ then absolute value of $a$  is equal to

For the point where the line intersects the curve, we have z = 0

So, $\frac{x-2}{3}=\frac{y-1}{2}=\frac{0-1}{-1}\Rightarrow x=5andy=3$

Putting these values in $x^2-y^2=a^2,$  we get $a^2=16i.e.,a=\pm 4$