Find the combined equation of pair of lines passing through (2, 3) and perpendicular to the lines \(3x + 2y - 1 = 0\) and \(x - 3y + 2 = 0\).

Answer: \(6x^2 - 7xy - 3y^2 - 3x + 32y - 45 = 0\)

Slope of \(3x + 2y - 1 = 0\) is \(-\frac{3}{2}\), so the required perpendicular line has slope \(\frac{2}{3}\). Through (2, 3): \(y - 3 = \frac{2}{3}(x - 2) \Rightarrow 2x - 3y + 5 = 0\).

Slope of \(x - 3y + 2 = 0\) is \(\frac{1}{3}\), so the perpendicular line has slope \(-3\). Through (2, 3): \(y - 3 = -3(x - 2) \Rightarrow 3x + y - 9 = 0\).

Combined equation: \((2x - 3y + 5)(3x + y - 9) = 0\)

\(6x^2 + 2xy - 18x - 9xy - 3y^2 + 27y + 15x + 5y - 45 = 0\)

\(\therefore 6x^2 - 7xy - 3y^2 - 3x + 32y - 45 = 0\).