Let $$A(h, k), B(1, 1)$$ and $$C(2, 1)$$ be the vertices of a right angled triangle with $$AC$$ as its hypotenuse. If the area of the triangle is 1, then the set of values which '$$k$$' can take is given by

Since $$AC$$ is the hypotenuse, the right angle must be at vertex $$B(1,1)$$. Hence the two sides through $$B$$, viz. $$\overline{AB}$$ and $$\overline{BC}$$, are perpendicular.

The slope of $$BC$$ is
$$m_{BC} = \frac{1-1}{2-1} = 0$$

A line perpendicular to a horizontal line (slope 0) is a vertical line whose slope is undefined. Therefore $$\overline{AB}$$ must be vertical, i.e. its $$x$$-coordinate is constant:
$$h = 1$$

Thus $$A$$ has coordinates $$(1,k)$$ and the two perpendicular legs of the right-angled triangle are
$$AB = |k-1|,\qquad BC = |2-1| = 1$$

The area of a right triangle is $$\dfrac12 \times(\text{product of the perpendicular legs})$$. Given that the area equals $$1$$, we have
$$\frac12 \times |k-1| \times 1 = 1$$
$$\Rightarrow\; |k-1| = 2$$

Solving the modulus equation:
$$k - 1 = 2 \;\;\text{or}\;\; k - 1 = -2$$
$$\Rightarrow\; k = 3 \;\;\text{or}\;\; k = -1$$

Hence the possible values of $$k$$ are $$\{-1,\,3\}$$.

Option C which is: $$\{-1, 3\}$$