After simplifying each case, use the discriminant to determine whether that quadratic has zero, one, or two real roots.

A root from the equation created for $x<k$ counts only if it is actually less than $k$. Apply the same check to roots from the $x\ge k$ branch.

To obtain three distinct solutions, consider which choice could make one branch contribute exactly one valid root and the other branch contribute two valid roots.

In Desmos, enter y=(x-2)^2 and y=abs(x-k). Add a slider for $k$.

Set the slider to each answer choice. The number of intersection points of the two graphs equals the number of distinct real solutions of $(x-2)^2=|x-k|$.

For the choice that appears to produce three intersections, select each intersection point to verify that the three $x$-coordinates are distinct. The graph can also confirm that the other choices each produce only two intersections.

The expression inside the absolute value is $x-k$. Therefore,

For $x<k$,

For $x\ge k$,

Any root found from a branch must also satisfy that branch's condition on $x$.

For the branch $x<k$, the discriminant is

For the branch $x\ge k$, the discriminant is

Thus, $k=\frac{7}{4}$ is notable because $D_1=0$, so the first branch has exactly one real root.

When $k=\frac{7}{4}$, the first branch becomes

which has the repeated root $x=\frac{3}{2}$. Since $\frac{3}{2}<\frac{7}{4}$, this root belongs to the first branch.

For the second branch,

Its discriminant is $2$, so it has two distinct roots:

Both are greater than $\frac{7}{4}$, so both belong to the second branch. Therefore, there are $1+2=3$ distinct real solutions.

For $k=1$ and $k=\frac{3}{2}$, $D_1<0$, while the second branch has two valid real roots, so there are only two solutions.

For $k=3$, $D_2<0$, while the first branch has two valid real roots, so there are again only two solutions. Therefore, the only choice that produces exactly three distinct real solutions is $\frac{7}{4}$.