1. Interpret the condition carefully

Since $AC$ is the hypotenuse of the right triangle $ABC$, the right angle must be at $B$.

So, $\angle ABC = 90^\circ$ which means vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are perpendicular.

Given: $A(h,k),\quad B(1,1),\quad C(2,1)$

2. Use perpendicularity at $B$

Compute the vectors: $\overrightarrow{BA}=(h-1,\,k-1)$ $\overrightarrow{BC}=(2-1,\,1-1)=(1,0)$

For perpendicular vectors, dot product is zero: $\overrightarrow{BA}\cdot \overrightarrow{BC}=0$ $(h-1, k-1)\cdot(1,0)=0$ $h-1=0$ $h=1$

So point $A$ must be of the form $A=(1,k)$

3. Use the area condition

Now the triangle has vertices: $A(1,k),\quad B(1,1),\quad C(2,1)$

Segment $BC$ is horizontal with length $BC=1$

Segment $AB$ is vertical with length $AB=|k-1|$

Since the right angle is at $B$, the area is $\text{Area}=\frac12 \cdot AB \cdot BC$ $1=\frac12\cdot |k-1|\cdot 1$ $|k-1|=2$

Hence, $k-1=\pm 2$ So, $k=3 \quad \text{or} \quad k=-1$

4. Set of possible values of $k$

Therefore, $\{k\}=\{-1,3\}$

5. Match with options

This corresponds to Option A.

6. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.