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

Given,

System of equations,

$x + 2y + z = 7$

$x + \alphaz = 11$

$2x - 3y + \betaz = \gamma$

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

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

$⇒ 7\alpha - 2\beta = 3$

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

Now finding,

{\Delta}_{3} = \left|\begin{cases}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & - 3 & \gamma\end{cases}\right|

Now equating, ${\Delta}_{3} = 0$ we get,

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

$⇒ \gamma = 28$

And {\Delta}_{1} = \left|\begin{cases}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & - 3 & \beta\end{cases}\right|

$⇒ {\Delta}_{1} = 21\alpha - 2\left(11\beta - \alpha\gamma\right) - 33$

$⇒ {\Delta}_{1} = 21\alpha - 22\beta + 2\alpha\gamma - 33$

Similarly, {\Delta}_{2} = \left|\begin{cases}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{cases}\right|

$⇒ {\Delta}_{2} = 11\beta - \alpha\gamma - 7$\left(\beta - 2\alpha\right) + \gamma - 22$</p><p>$⇒ {\Delta}_{2} = 14\alpha + 4\beta + \gamma - \alpha\gamma - 22$</p><p>\left(P\right)$ If $\beta = \dfrac{1}{2}$\left(7\alpha - 3\right)$and$\gamma = 28$</p><p>$\Delta = 0, {\Delta}_{1} = 0, {\Delta}_{2} = 0, {\Delta}_{3} = 0$</p><p>Infinitely many solutions</p><p>$x = 11, y = - 2$and$z = 0$will satisfy all the three given equations, so it is a solution.</p><p>$\left(Q\right)$If$\beta = \dfrac{1}{2}$\left(7\alpha - 3\right)$ and $\gamma \ne 28$ then

$\Delta = 0$, but ${\Delta}_{3} \ne 0$ so no solution

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

$\Delta \ne 0, {\Delta}_{3} \ne 0$ so a unique solution

\left(S\right)$If$\beta \ne \dfrac{1}{2}\left(7\alpha - 3\right), \alpha = 1, \gamma = 28$</p><p>$\Delta \ne 0, {\Delta}_{3} = 0, {\Delta}_{1} \ne 0, {\Delta}_{2} \ne 0$, so a unique solution</p><p>$x = 11, y = - 2$and$z = 0 will satisfy all the three equations

Option A is correct.