Find $k$ if $(1, 7), (2, 4)$ and $(k, 5)$ form a right triangle.

(A-D) Let the vertices be $A(1, 7), B(2, 4)$,and $C(k, 5)$.
Calculate the squared distances between the points:
$AB^2 = (1-2)^2 + (7-4)^2 = (-1)^2 + 3^2 = 1 + 9 = 10$
$BC^2 = (2-k)^2 + (4-5)^2 = 4 - 4k + k^2 + 1 = k^2 - 4k + 5$
$AC^2 = (1-k)^2 + (7-5)^2 = 1 - 2k + k^2 + 4 = k^2 - 2k + 5$
For a right triangle,one of the angles must be $90^{\circ}$:
Case $1$: $\angle A = 90^{\circ}$. Then $BC^2 = AB^2 + AC^2$.
$k^2 - 4k + 5 = 10 + k^2 - 2k + 5$
$-2k = 10 \implies k = -5$
Case $2$: $\angle B = 90^{\circ}$. Then $AC^2 = AB^2 + BC^2$.
$k^2 - 2k + 5 = 10 + k^2 - 4k + 5$
$2k = 10 \implies k = 5$
Case $3$: $\angle C = 90^{\circ}$. Then $AB^2 = BC^2 + AC^2$.
$10 = (k^2 - 4k + 5) + (k^2 - 2k + 5)$
$10 = 2k^2 - 6k + 10$
$2k^2 - 6k = 0 \implies 2k(k - 3) = 0$
$k = 0$ or $k = 3$
Thus,the possible values of $k$ are $-5, 5, 0, 3$.