Explanation

(b)
Let $S_{1} + \lambda\ S_{2} = 0$
Since, it passes through (1, 1), then
$(1 + 1 + 3 + 7 + 2p - 5) + \lambda\left( 1 + 1 + 2 + 2 - p^{2} \right) = 0$
$\Rightarrow \ \ \ \lambda = - \frac{7 + 2p}{6 - p^{2}}$
But $p^{2} \neq 6\ \ \Rightarrow \ \ \ p \neq \pm \sqrt{6}$
But the other circle $x^{2} + y^{2} + 2x + 2y - 6 = 0$ at $p = \pm \sqrt{6}$ also satisfy the point
(1, 1)
So, $p = \pm \sqrt{6}$ is valid
Now, $\lambda \neq - 1\ \ \Rightarrow \ \ \frac{7 + 2p}{6 - p^{2}} \neq 1\ \$
$\Rightarrow \ \ \ 7 + 2p \neq 6 - p^{2}$
$\Rightarrow \ \ \ p^{2} + 2p + 1 \neq 0\ \ \Rightarrow \ \ \ p \neq - 1$