Ans.

(1)

$\frac{x – 3}{2} = \frac{y – 10}{1} = \frac{–2(2 \times 3 + 10 – 6)}{4 + 1} = –4$

$\Rightarrow x = –5, y = 6$

$\therefore$    A'(–5, 6) and B(7, 2)

$\therefore$    Equation of line A'B is

$\Rightarrow y – 6 = \frac{2 – 6}{7 + 5} (x + 5)$

$\Rightarrow y – 6 = – \frac{1}{3} (x + 5)$

$\Rightarrow 3y – 18 = –x – 5 \Rightarrow x + 3y = 13$

and 2x + y = 6 (Given line).

On solving, we get y = 4, x = 1

$\therefore Q \equiv (1, 4)$

Equation of line AQ is

$y – 10 = \frac{10 – 4}{3 – 1} (x – 3) \Rightarrow y – 10 = 3(x – 3)$

$\Rightarrow y – 10 = 3x – 9 \Rightarrow 3x – y + 1 = 0$

On comparing with given equation ax + by + 1 = 0, we get a = 3, b = –1

Hence, $a^2 + b^2 + 3ab = 9 + 1 + 3\left(3\right)(–1) = 9 + 1 – 9 = 1$.