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 | | --- | --- | | (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 |

$x+2y+z=7$

$x+\alpha z=11$

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

$\text{Match } (P),(Q),(R),(S) \text{ with } (1)-(5)$

$\text{using the conditions on } \beta \text{ and } \gamma$

$\Delta \neq 0 \Rightarrow \text{unique solution}$

$\Delta = 0 \Rightarrow \text{no solution or infinitely many}$

$\text{Test cases } (P),(Q),(R),(S)$

Δ = 1 2 1 1 0 α 2 -3 β

$= 1(3\alpha) - 2(\beta - 2\alpha) + 1(-3)$

$= 3\alpha - 2\beta + 4\alpha - 3 = 7\alpha - 2\beta - 3$

$\Delta = 0 \Leftrightarrow 7\alpha - 2\beta - 3 = 0$

$\Leftrightarrow \beta = \dfrac{7\alpha-3}{2}$

$R_3 = -\dfrac{3}{2}R_1 + \dfrac{7}{2}R_2$

$\text{Consistent} \Leftrightarrow \gamma = -\dfrac{3}{2}(7) + \dfrac{7}{2}(11) = 28$

$(P):\ \gamma=28 \Rightarrow \text{consistent},\ \Delta=0$

$\Rightarrow \text{infinitely many solutions } \to (3)$

$(Q):\ \gamma\neq 28 \Rightarrow \text{inconsistent}$

$\Rightarrow \text{no solution } \to (2)$

$(R):\ \alpha=1 \Rightarrow \Delta = 4-2\beta \neq 0$

$\Rightarrow \text{unique solution } \to (1)$

$(S):\ \alpha=1,\ \beta\neq 2,\ \gamma=28 \Rightarrow \Delta\neq 0$

$(11,-2,0):\ 11+2(-2)=7,\quad 11=11,\quad 22+6=28\ \checkmark$

$\Rightarrow \text{solution } (11,-2,0) \to (4)$

$(P)\to(3),\quad (Q)\to(2),\quad (R)\to(1),\quad (S)\to(4)$

$\boxed{\text{Answer: option (1)}}$