Explanation
a, b, c are in A.P
\begin{align*}
2b &= a + c \\
a - 2b + c &= 0
\end{align*}
Comparing with ax+by+c=0, we get:
x=1,y=−2
∴P(1,−2)
For infinite solutions of the system:
Δ=Δx=Δy=Δz=0
\begin{align*}
\Delta &= \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 0 \\
&\Rightarrow 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0 \\
&\Rightarrow 15 - 2\alpha - 6 + \alpha - 1 = 0 \\
&\Rightarrow 8 - \alpha = 0 \Rightarrow \alpha = 8
\end{align*}
\begin{align*}
\Delta_z &= \begin{vmatrix} 1 & 1 & 6 \\ 2 & 5 & \beta \\ 1 & 2 & 4 \end{vmatrix} = 0 \\
&\Rightarrow 1(20 - 2\beta) - 1(8 - \beta) + 6(4 - 5) = 0 \\
&\Rightarrow 20 - 2\beta - 8 + \beta - 6 = 0 \\
&\Rightarrow \beta = 6
\end{align*}
∴Q(8,6) \\
\begin{align*}
(PQ)^2 &= (8 - 1)^2 + (6 - (-2))^2 \\
&= 7^2 + 8^2 \\
&= 49 + 64 \\
&= 113
\end{align*}
