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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-$II$
| 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 |
Then the system has
$(\mathrm{P}) \rightarrow(3)(\mathrm{Q}) \rightarrow(2)(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(4)$
$(P) \rightarrow (3) (Q) \rightarrow (2) (R) \rightarrow (5) (S) \rightarrow (4)$
$(P) \rightarrow (2) (Q) \rightarrow (1) (R) \rightarrow (4) (S) \rightarrow (5)$
$(\mathrm{P}) \rightarrow(2)(\mathrm{Q}) \rightarrow(1)(\mathrm{R}) \rightarrow(1)(\mathrm{S}) \rightarrow(3)$
Solution
$\begin{aligned} & \text { Given } x+2 y+z=7 \\ & x+\alpha z=11 \\ & 2 x-3 y+\beta z=\gamma \\ & \text { Now, } \Delta=\left|\begin{array}{ccc}1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta\end{array}\right|=7 \alpha-2 \beta-3 \\ & \therefore \text { if } \beta=\frac{1}{2}(7 \alpha-3) \\ & \Rightarrow \Delta=0 \\ & \text { Now, } \Delta_{\mathrm{x}}=\left|\begin{array}{ccc}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta\end{array}\right| \\ & =21 \alpha-22 \beta+2 \alpha \gamma-33 \\ & \therefore \text { if } \gamma=28 \\ & \Rightarrow \Delta_{\mathrm{x}}=0 \\ & \end{aligned}$
$\begin{aligned} & \Delta_y=\left|\begin{array}{ccc}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{array}\right| \\ & \Delta_y=4 \beta+14 \alpha-\alpha \gamma+\gamma-22 \\ & \therefore \text { if } \gamma=28 \\ & \Rightarrow \Delta_y=0 \\ & \text { Now, } \Delta_z=\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right|=56-2 \gamma \\ & \text { If } \gamma=28 \\ & \Rightarrow \Delta_z=0 \\ & \therefore \text { if } \gamma=28 \text { and } \beta=\frac{1}{2}(7 \alpha-3)\end{aligned}$
$\therefore \text { if } \gamma=28 \text { and } \beta=\frac{1}{2}(7 \alpha-3)$
$\Rightarrow \text { system has infinite solution }$
$\text { and if } \gamma \neq 28$
$\Rightarrow \text { system has no solution }$
$\Rightarrow P \rightarrow(3) ; Q \rightarrow(2)$
$\text { Now if } \beta \neq \frac{1}{2}(7 \alpha-3)$
$\Rightarrow \Delta \neq 0$
$\text { and for } \alpha=1 \text { clearly }$
$y=-2 \text { is always be the solution }$
$\therefore \text { if } \gamma \neq 28$
$\text { System has a unique solution }$
$\text { if } \gamma=28$
$\Rightarrow x=11, y=-2 \text { and } z=0 \text { will be one of the solution }$
$\therefore R \rightarrow 1 ; S \rightarrow 4$
$\therefore \text { option 'A' is correct }$
