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(B) Since $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $a - 2b + c = 0$.This means the line $ax + by + c = 0$ always passes through the

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$113$

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(B) Since $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $a - 2b + c = 0$.This means the line $ax + by + c = 0$ always passes through the fixed point $P(1, -2)$.For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.$D = \begin{vmatrix} 1 & 1 & 1 \ 2 & 5 & \alpha \ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$.$15 - 2\alpha - 6 + \alpha - 1 = 0 \Rightarrow 8 - \alpha = 0 \Rightarrow \alpha = 8$.Now,for infinite solutions,$D_1 = 0$ where $D_1 = \begin{vmatrix} 6 & 1 & 1 \ \beta & 5 & 8 \ 4 & 2 & 3 \end{vmatrix} = 0$.$6(15 - 16) - 1(3\beta - 32) + 1(2\beta - 20) = 0$.$-6 - 3\beta + 32 + 2\beta - 20 = 0 \Rightarrow -\beta + 6 = 0 \Rightarrow \beta = 6$.Thus,$Q = (8, 6)$.The distance $PQ = \sqrt{(8 - 1)^2 + (6 - (-2))^2} = \sqrt{7^2 + 8^2} = \sqrt{49 + 64} = \sqrt{113}$.Therefore,$(PQ)^2 = 113$.

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