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

$x+2y+z=7$

$x+\alpha z=11$

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

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

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

The correct option is:

• Correct answer
  A

  $\left(\mathrm{P}\right)\to \left(3\right), \left(\mathrm{Q}\right)\to \left(2\right), \left(\mathrm{R}\right)\to \left(1\right), \left(\mathrm{S}\right)\to \left(4\right)$
• B

  $\left(\mathrm{P}\right)\to \left(3\right), \left(\mathrm{Q}\right)\to \left(2\right), \left(\mathrm{R}\right)\to \left(5\right), \left(\mathrm{S}\right)\to \left(4\right)$
• C

  $\left(\mathrm{P}\right)\to \left(2\right), \left(\mathrm{Q}\right)\to \left(1\right), \left(\mathrm{R}\right)\to \left(4\right), \left(\mathrm{S}\right)\to \left(5\right)$
• D

  $\left(\mathrm{P}\right)\to \left(2\right), \left(\mathrm{Q}\right)\to \left(1\right), \left(\mathrm{R}\right)\to \left(1\right), \left(\mathrm{S}\right)\to \left(3\right)$

Hence, the correct answer is option: 'A'.