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)?

A. $$\frac{{37}}{{35}}$$

B. $$\frac{5}{7}$$

C. $$\frac{{31}}{{41}}$$

D. $$\frac{{41}}{{31}}$$

Answer: Option C

Solution(By Examveda Team)

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)
$$\eqalign{ & \frac{{{a^2} + {b^2} - ab}}{{{a^2} - {b^2} + ab}} \cr & = \frac{{36 + 1 - 6}}{{36 - 1 + 6}} \cr & = \frac{{31}}{{41}} \cr} $$