Let \(C(k,7) = (x_1,y_1), A(-5,1) = (x_2,y_2), B(3, 5) = (x_3, y_3)\)
Since points are equidistant, so
AC = BC
\(AC^2 = BC^2\)
Using distance formula,
\((x_3-x_1)^2 + (y_3-y_1)^2 = (x_2-x_1)^2+(y_2-y_1)^2\)
\(\Rightarrow (3-k)^2+(5-7)^2 = (-5-k)^2+(1-7)^2\)
\(\Rightarrow 9 - 6k + k^2 + (-2)^2 = 25 + 10k + k^2 + (-6)^2\)
\(\Rightarrow 9-6k+k^2+4=25+10k+k^2+36\)
\(\Rightarrow 13-6k=61+10k\)
\(\Rightarrow -6k-10k =61-13\)
\(\Rightarrow -16k = 48\)
\(\Rightarrow k = \cfrac{48}{-16}\)
\(\Rightarrow k = -3\)