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

Given,

System of equations,

$x+2y+z=7$

$x+\alpha z=11$

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

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

$\Rightarrow 3\alpha -2\left(\beta -2\alpha \right)-3=0$

$\Rightarrow 7\alpha -2\beta =3$

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

Now finding,

${\Delta }_3=\left|\begin{array}{ccc}1& 2& 7\\ 1& 0& 11\\ 2& -3& \gamma \end{array}\right|$

Now equating, ${\Delta }_3=0$ we get,

$\Rightarrow 33-2\left(\gamma -22\right)+7\left(-3\right)=0$

$\Rightarrow \gamma =28$

And ${\Delta }_1=\left|\begin{array}{ccc}7& 2& 1\\ 11& 0& \alpha \\ \gamma & -3& \beta \end{array}\right|$

 $\Rightarrow {\Delta }_1=21\alpha -2\left(11\beta -\alpha \gamma \right)-33$

$\Rightarrow {\Delta }_1=21\alpha -22\beta +2\alpha \gamma -33$

Similarly, ${\Delta }_2=\left|\begin{array}{ccc}1& 7& 1\\ 1& 11& \alpha \\ 2& \gamma & \beta \end{array}\right|$

$\Rightarrow {\Delta }_2=11\beta -\alpha \gamma -7\left(\beta -2\alpha \right)+\gamma -22$

$\Rightarrow {\Delta }_2=14\alpha +4\beta +\gamma -\alpha \gamma -22$

$\left(\mathrm{P}\right)$ If $\beta =\frac{1}{2}\left(7\alpha -3\right)$ and $\gamma =28$

$\Delta =0, {\Delta }_1=0, {\Delta }_2=0, {\Delta }_3=0$

Infinitely many solutions

$x=11, y=-2$ and $z=0$ will satisfy all the three given equations, so it is a solution.

$\left(\mathrm{Q}\right)$ If $\beta =\frac{1}{2}\left(7\alpha -3\right)$ and $\gamma \ne 28$ then

$\Delta =0$, but ${\Delta }_3\ne 0$ so no solution

$\left(\mathrm{R}\right)$ If $\beta \ne \frac{1}{2}\left(7\alpha -3\right), \alpha =1$ and $\gamma \ne 28$

$\Delta \ne 0, {\Delta }_3\ne 0$ so a unique solution

$\left(\mathrm{S}\right)$ If $\beta \ne \frac{1}{2}\left(7\alpha -3\right), \alpha =1, \gamma =28$

$\Delta \ne 0, {\Delta }_3=0, {\Delta }_1\ne 0, {\Delta }_2\ne 0$, so a unique solution

$x=11, y=-2$ and $z=0$ will satisfy all the three equations

Option A is correct.