The provided line equation is \( 4x - y - 2 = 0 \). Point \( P \) is equidistant from \( A(-5, 6) \) and \( B(3, 2) \), meaning \( PA = PB \). This implies \( PA^2 = PB^2 \).
Step 1: Apply the distance formula.
The distance from \( P(a, b) \) to \( A(-5, 6) \) is \( \sqrt{(a + 5)^2 + (b - 6)^2} \), and to \( B(3, 2) \) is \( \sqrt{(a - 3)^2 + (b - 2)^2} \). Setting these equal and squaring both sides yields:
\[
(a + 5)^2 + (b - 6)^2 = (a - 3)^2 + (b - 2)^2.
\]
Expanding and simplifying:
\[
a^2 + 10a + 25 + b^2 - 12b + 36 = a^2 - 6a + 9 + b^2 - 4b + 4.
\]
Combining like terms results in:
\[
16a - 8b + 48 = 0.
\]
Step 2: Solve the system of equations.
From the line equation \( 4x - y - 2 = 0 \), we express \( b \) in terms of \( a \) as \( b = 4a - 2 \). Substituting this into the simplified distance equation:
\[
16a - 8(4a - 2) + 48 = 0.
\]
Simplifying this equation:
\[
16a - 32a + 16 + 48 = 0,
\]
\[
-16a + 64 = 0.
\]
Solving for \( a \), we get \( a = 4 \). Substituting \( a = 4 \) back into the equation for \( b \):
\[
b = 4(4) - 2 = 14.
\]
Final Answer:
The coordinates of point \( P \) are:
\[
\boxed{(4, 14)}.
\]