Step 1 :Set the two equations equal to each other: \(x + 12 = 2 - 4x\).
Step 2 :Solve for x: \(5x = -10\) so \(x = -2\).
Step 3 :Substitute \(x = -2\) into either of the original equations to find y. Using the first equation: \(y = -2 + 12 = 10\).
Step 4 :So, the coordinates of point P are \(\boxed{(-2, 10)}\).
Step 5 :The gradient of line AB is \(\frac{8q-(-q)}{5p-2p} = \frac{9q}{3p} = \frac{3q}{p}\).
Step 6 :Since the line perpendicular to AB has gradient \(-\frac{2}{3}\), and the product of the gradients of two perpendicular lines is -1, we have \(-\frac{2}{3} = -\frac{1}{\frac{3q}{p}}\).
Step 7 :Solving for p, we get \(p = \frac{3q}{2}\).