Possible Right Triangles

Let the vertices of the triangle formed be (3, 7), (6, 2), and (2, k).

Case 1: (3, 7) is the right angled vertex. Form the vectors that corresponding to the legs of the right triangle: $\langle 3, -5 \rangle$ and $\langle -1, k - 7 \rangle$ . Their dot product must be zero, so this gives

$-3 - 5(k - 7) = 0 \longrightarrow k = 32/5$

Case 2: (6, 2) is the right angled vertex. Form the vectors that corresponding to the legs of the right triangle: $\langle -3, 5 \rangle$ and $\langle -4, k - 2 \rangle$ . Again, the dot product must be zero, so this gives

$12 + 5(k - 2) = 0 \longrightarrow k = -2/5$

Case 3: (2, k) is the right angled vertex. Form the vectors that corresponding to the legs of the right triangle: $\langle 1, 7 - k \rangle$ and $\langle 4, 2 - k \rangle$ . Setting the dot product equal to zero yields

$4 + (7 - k)(2 - k) = 0 \longrightarrow k = 3, k = 6$

This is gives 4 distinct values of k for which we could have a right triangle.

Let (3,7)(6,2) be the diameter of a circle C. The perpendicular from the center of the C to x=2 is shorter than its radius. So the semicircle on the diameter will cut x=2 at two points giving two right angles. We can have right angle at (3,7) and (6,2) also. So there are four right angles.
If the perpendicular distance of center of C from x=2 is just its radius, x=2 will be the tangent and meet x=2 only at one point in place of two we now have. Needless to say that there will be no right angle if distance is more than the radius.

The choices answer the question already! there are 3 cases so the answer is 3 values or more. You have to reformulate it.