(a) \(x^{2} + mx + 11 = 0\)
Roots are \(\alpha\) and \(\beta\).
\(\therefore \alpha + \beta = \frac{-b}{a} = - m & \alpha \beta = \frac{c}{a} = 11\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)
\(27 = (-m)^{2} - 2(11)\)
\(m^{2} = 22 + 27 = 49\)
\(m = \pm 7\)
(b) Equation of line : \(2x + 3y = 1 i.e. y = \frac{1 - 2x}{3}\)
Substitute for y in the equation of the circle:
\(2x^{2} + 2(\frac{1 - 2x}{3})^{2} + 4x + 9(\frac{1 - 2x}{3}) - 9 = 0\)
\(2x^{2} + 2(\frac{1 - 4x + 4x^{2}}{9} + 4x + 3(1 - 2x) - 9 = 0\)
\(18x^{2} + 2 - 8x + 8x^{2} + 36x + 27 - 54x - 81 = 0\)
\(26x^{2} - 26x - 52 = 0\)
\(x^{2} - x - 2 = 0\)
\((x - 2)(x + 1) = 0\)
\(x = \text{2 or -1}\)
Substitute for x in \(y = \frac{1 - 2x}{3}\); when x = 2,
\(y = \frac{1 - 2(2)}{3} = \frac{-3}{3} = -1\)
Coordinate = \((2, -1)\).
When x = -1,
\(y = \frac{1 - 2(-1)}{3} = \frac{3}{3} = 1\)
Coordinate = \((-1, 1)\)
Since Q lies in the fourth quadrant, we have P(-1, 1) and Q(2, -1).