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(A) For the system to have at least one solution,the determinant $\Delta$ must be non-zero,or if $\Delta = 0$,then $\Delta_1 = \Delta_2 = \Delta_3 = 0

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$A, C, D$

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(A) For the system to have at least one solution,the determinant $\Delta$ must be non-zero,or if $\Delta = 0$,then $\Delta_1 = \Delta_2 = \Delta_3 = 0$.For the given system:$\Delta = \begin{vmatrix} -1 & 2 & 5 \ 2 & -4 & 3 \ 1 & -2 & 2 \end{vmatrix} = -1(-8+6) - 2(4-3) + 5(-4+4) = 2 - 2 + 0 = 0$.For the system to have a solution,we must have $\Delta_1 = \Delta_2 = \Delta_3 = 0$.Calculating $\Delta_1 = \begin{vmatrix} b_1 & 2 & 5 \ b_2 & -4 & 3 \ b_3 & -2 & 2 \end{vmatrix} = b_1(-8+6) - 2(2b_2-3b_3) + 5(-2b_2+4b_3) = -2b_1 - 4b_2 + 6b_3 - 10b_2 + 20b_3 = -2b_1 - 14b_2 + 26b_3 = 0$.This simplifies to $b_1 + 7b_2 - 13b_3 = 0$. Thus,$S = \{ [b_1, b_2, b_3]^T : b_1 + 7b_2 - 13b_3 = 0 \}$.For each option,we check if the system has a solution for all $[b_1, b_2, b_3]^T \in S$. If $\Delta 
eq 0$,it always has a unique solution.$(A)$ $\Delta = \begin{vmatrix} 1 & 2 & 3 \ 0 & 4 & 5 \ 1 & 2 & 6 \end{vmatrix} = 1(24-10) - 2(0-5) + 3(0-4) = 14 + 10 - 12 = 12 
eq 0$. Solution exists.$(B)$ $\Delta = \begin{vmatrix} 1 & 1 & 3 \ 5 & 2 & 6 \ -2 & -1 & -3 \end{vmatrix} = 0$. Checking $\Delta_1$: $\begin{vmatrix} b_1 & 1 & 3 \ b_2 & 2 & 6 \ b_3 & -1 & -3 \end{vmatrix} = b_1(0) - 1(-3b_2-6b_3) + 3(-b_2-2b_3) = 3b_2+6b_3-3b_2-6b_3 = 0$. Since $\Delta_1 = \Delta_2 = \Delta_3 = 0$,it has solutions,but we must check if it holds for all $S$. Actually,for $(B)$,the equations are dependent,and it has solutions for all $b_i$.$(C)$ $\Delta = 0$ and $\Delta_1 = \Delta_2 = \Delta_3 = 0$. Solution exists.$(D)$ $\Delta = \begin{vmatrix} 1 & 2 & 5 \ 2 & 0 & 3 \ 1 & 4 & -5 \end{vmatrix} = 1(0-12) - 2(-10-3) + 5(8-0) = -12 + 26 + 40 = 54 
eq 0$. Solution exists.Thus,$(A), (B), (C), (D)$ all have solutions.

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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