Let α, β, and γ be real numbers. Consider the following system of linear equations

x + 2y + z = 7

x + αz = 11

2x − 3y + βz = γ

Match each entry in List-I to the correct entries in List-II

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

The correct option is (P) → (3), (Q) → (2), (R) → (1), (S) → (4).

Consider the given system of linear equations:

1. $x+2y+z=7$
2. $x+0y+\alpha z=11$
3. $2x-3y+\beta z=\gamma$

Let us calculate the coefficient determinant $D$ of the system:

$D=\left|\begin{array}{ccc}1& 2& 1\\ 1& 0& \alpha \\ 2& -3& \beta \end{array}\right|$

Expanding along the second row:

$D=-1(2\beta +3)-\alpha (-3-4)=-2\beta -3+7\alpha =7\alpha -2\beta -3$

Setting $D=0$ gives:

$7\alpha -2\beta -3=0\Rightarrow \beta =\frac{1}{2}(7\alpha -3)$

Now, let us calculate $D_z$ by replacing the third column with the constants vector:

$D_z=\left|\begin{array}{ccc}1& 2& 7\\ 1& 0& 11\\ 2& -3& \gamma \end{array}\right|$

Expanding along the first row:

$D_z=1(0+33)-2(\gamma -22)+7(-3-0)=33-2\gamma +44-21=56-2\gamma$

Setting $D_z=0$ gives:

$56-2\gamma =0\Rightarrow \gamma =28$

Analysis of Cases:

Case (P): If $\beta =\frac{1}{2}(7\alpha -3)$ and $\gamma =28$
Here, $D=0$ and $D_x=D_y=D_z=0$. Therefore, the system has infinitely many solutions.
Thus, (P) → (3).

Case (Q): If $\beta =\frac{1}{2}(7\alpha -3)$ and $\gamma \ne 28$
Here, $D=0$ but $D_z\ne 0$. Therefore, the system has no solution.
Thus, (Q) → (2).

Case (R): If $\beta \ne \frac{1}{2}(7\alpha -3)$ where $\alpha =1$ and $\gamma \ne 28$
Here, $D\ne 0$. Since the coefficient determinant is non-zero, the system has a unique solution.
Thus, (R) → (1).

Case (S): If $\beta \ne \frac{1}{2}(7\alpha -3)$ where $\alpha =1$ and $\gamma =28$
Let us test the given solution $x=11,y=-2,z=0$ from List-II (4):
Equation 1: $11+2(-2)+0=11-4=7$ (Satisfied)
Equation 2 (with $\alpha =1$): $11+1\left(0\right)=11$ (Satisfied)
Equation 3 (with $\gamma =28$): $2\left(11\right)-3(-2)+\beta \left(0\right)=22+6=28$ (Satisfied)
Since $x=11,y=-2,z=0$ satisfies all three equations, it is a valid solution.
Thus, (S) → (4).