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

Write the three equations in the standard form

$$\begin{aligned} x+2y+z &= 7 \hspace{20pt} -(1)\\ x+\alpha z &= 11 \hspace{20pt} -(2)\\ 2x-3y+\beta z &= \gamma \hspace{20pt} -(3) \end{aligned}$$

The coefficient matrix is

$$A=\begin{bmatrix} 1 & 2 & 1\\ 1 & 0 & \alpha\\ 2 & -3 & \beta \end{bmatrix}$$

The determinant of $$A$$ decides whether a unique solution exists.

$$\begin{aligned} \det(A) &= \begin{vmatrix} 1 & 2 & 1\\ 1 & 0 & \alpha\\ 2 & -3 & \beta \end{vmatrix}\\[2pt] &=1\,(0\cdot\beta-\alpha(-3))- 2\,(1\cdot\beta-\alpha\cdot2)+ 1\,(1\cdot(-3)-0\cdot2)\\[2pt] &=1\,(3\alpha)-2\,(\beta-2\alpha)+(-3)\\[2pt] &=3\alpha-2\beta+4\alpha-3\\ &=7\alpha-2\beta-3 \end{aligned}$$

Thus

$$\det(A)=0 \;\Longleftrightarrow\; 2\beta=7\alpha-3 \quad -(4)$$

When $$\det(A)\neq0$$ the system has a unique solution; when $$\det(A)=0$$ we must compare the ranks of $$A$$ and the augmented matrix $$[A|B]$$ to decide between “no solution” and “infinitely many solutions”.

Next, express row (3) of $$[A|B]$$ as a linear combination of rows (1) and (2).
Take numbers $$p,q$$ so that

$$p(1,2,1)+q(1,0,\alpha)=(2,-3,\beta)$$

Equating components gives

$$p+q=2,\;2p=-3\;\Longrightarrow\;p=-\tfrac32,\;q=\tfrac72$$

With this $$z$$-coefficient becomes

$$p\cdot1+q\cdot\alpha=-\tfrac32+\tfrac72\alpha=\beta$$

which is exactly condition (4). The corresponding constant term is

$$p\cdot7+q\cdot11=-\tfrac32\cdot7+\tfrac72\cdot11 =-\tfrac{21}{2}+\tfrac{77}{2}=28$$

Hence, when (4) holds:
  • If $$\gamma=28$$, row (3) is a linear combination of rows (1) and (2) ⇒ ranks are equal (both 2 < 3) ⇒ infinitely many solutions.
  • If $$\gamma\neq28$$, the augmented matrix has rank 3 while $$\operatorname{rank}(A)=2$$ ⇒ no solution.

Now fix $$\alpha=1$$ as given in cases (R) and (S). From (4) we get the special value

$$2\beta=7(1)-3\;\Longrightarrow\;\beta=2$$

Therefore, for $$\alpha=1$$:

$$\det(A)=7(1)-2\beta-3=4-2\beta$$

If $$\beta\neq2$$ then $$\det(A)\neq0$$, so the system has a unique solution irrespective of $$\gamma$$. In particular, substituting $$\alpha=1$$ and solving quickly:

From (2): $$x=11-z$$. Substituting into (1): $$11-z+2y+z=7\Rightarrow y=-2$$.
Using (3): $$2(11-z)-3(-2)+\beta z=\gamma$$ gives $$28+(\beta-2)z=\gamma$$.

• For $$\gamma=28$$ we can take $$z=0$$, leading to the specific solution $$x=11,\;y=-2,\;z=0$$ (unique because $$\beta\neq2$$ keeps $$\det(A)\neq0$$).
• For $$\gamma\neq28$$, solving $$z=\dfrac{\gamma-28}{\beta-2}$$ yields a single real triple, again a unique solution.

We are now ready to match.

Case P: $$\beta=\tfrac12(7\alpha-3),\;\gamma=28$$ ⇒ $$\det(A)=0$$ and ranks equal ⇒ infinitely many solutions ⇒ List-II (3).

Case Q: $$\beta=\tfrac12(7\alpha-3),\;\gamma\neq28$$ ⇒ $$\det(A)=0$$ but ranks unequal ⇒ no solution ⇒ List-II (2).

Case R: $$\alpha=1,\;\beta\neq2,\;\gamma\neq28$$ ⇒ $$\det(A)\neq0$$ ⇒ unique solution ⇒ List-II (1).

Case S: $$\alpha=1,\;\beta\neq2,\;\gamma=28$$ gives the concrete solution $$x=11,\;y=-2,\;z=0$$ (and the solution is unique) ⇒ List-II (4).

The only option that lists these four pairings is:

Option A which is: (P) → (3), (Q) → (2), (R) → (1), (S) → (4).