Let $\alpha , \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations

$x + 2 y + z = 7$

$x + \alpha z = 11$

$2 x - 3 y + \beta z = \gamma$

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

\textbf{List-I}

(P) If $\beta = \frac { 1 } { 2 } ( 7 \alpha - 3 )$ and $\gamma = 28$, then the system has

(Q) If $\beta = \frac { 1 } { 2 } ( 7 \alpha - 3 )$ and $\gamma \neq 28$, then the system has

(R) If $\beta \neq \frac { 1 } { 2 } ( 7 \alpha - 3 )$ where $\alpha = 1$ and $\gamma \neq 28$, then the system has

(S) If $\beta \neq \frac { 1 } { 2 } ( 7 \alpha - 3 )$ where $\alpha = 1$ and $\gamma = 28$, then the system has

\textbf{List-II}

(1) a unique solution

(2) no solution

(3) infinitely many solutions

(4) $x = 11 , y = - 2$ and $z = 0$ as a solution

(5) $x = - 15 , y = 4$ and $z = 0$ as a solution

The correct option is:

(A) $( P ) \rightarrow ( 3 ) \quad ( Q ) \rightarrow ( 2 ) \quad ( R ) \rightarrow ( 1 ) \quad ( S ) \rightarrow ( 4 )$

(B) $( P ) \rightarrow ( 3 ) \quad ( Q ) \rightarrow ( 2 ) \quad ( R ) \rightarrow ( 5 ) \quad ( S ) \rightarrow ( 4 )$

(C) $( P ) \rightarrow ( 2 ) \quad ( Q ) \rightarrow ( 1 ) \quad ( R ) \rightarrow ( 4 ) \quad ( S ) \rightarrow ( 5 )$

(D) $( P ) \rightarrow ( 2 ) \quad ( Q ) \rightarrow ( 1 ) \quad ( R ) \rightarrow ( 1 ) \quad ( S ) \rightarrow ( 3 )$