The graphs of the equations 3x - 20y - 2 = 0 and 11x - 5y + 61 = 0 intersect at P(a, b). What is the value of (a2 + b2 - ab)(a2 - b2 + ab)?

Answer: Option C

It intersect at point P(a, b) = (x, y)
3x - 20y = 2
11x - 5y = -61
44x - 20y = -244
3x - 20y = 2
$$\overline{41\text{x}=-246}$$
x = -6
y = -1
P(a, b) = (-6, -1)
$$\begin{array}{rl}& \frac{a^2+b^2-ab}{a^2-b^2+ab}\\ & =\frac{36+1-6}{36-1+6}\\ & =\frac{31}{41}\end{array}$$