Question:

Step 1: Write down the system of equations and calculate the determinant of the coefficient matrix $D$. The given system of linear equations is: $$ x+y+z=6 \\ \alpha x+\beta y+7z=3 \\ x+2y+3z=14 $$ The coefficient matrix is $A = \begin{pmatrix} 1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3 \end{pmatrix}$. The determinant of the coefficient matrix is $D$: $$ D = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3 \end{vmatrix} $$ Expanding the determinant along the first row: $$ D = 1(\beta \cdot 3 - 7 \cdot 2) - 1(\alpha \cdot 3 - 7 \cdot 1) + 1(\alpha \cdot 2 - \beta \cdot 1) \\ D = (3\beta - 14) - (3\alpha - 7) + (2\alpha - \beta) \\ D = 3\beta - 14 - 3\alpha + 7 + 2\alpha - \beta \\ D = 2\beta - \alpha - 7 $$ Step 2: Calculate the determinants $D_x, D_y, D_z$. To calculate $D_x$, replace the first column of $D$ with the constant terms: $$ D_x = \begin{vmatrix} 6 & 1 & 1 \\ 3 & \beta & 7 \\ 14 & 2 & 3 \end{vmatrix} \\ D_x = 6(3\beta - 14) - 1(9 - 98) + 1(6 - 14\beta) \\ D_x = 18\beta - 84 - (-89) + 6 - 14\beta \\ D_x = 4\beta + 11 $$ To calculate $D_y$, replace the second column of $D$ with the constant terms: $$ D_y = \begin{vmatrix} 1 & 6 & 1 \\ \alpha & 3 & 7 \\ 1 & 14 & 3 \end{vmatrix} \\ D_y = 1(9 - 98) - 6(3\alpha - 7) + 1(14\alpha - 3) \\ D_y = -89 - 18\alpha + 42 + 14\alpha - 3 \\ D_y = -4\alpha - 50 $$ To calculate $D_z$, replace the third column of $D$ with the constant terms: $$ D_z = \begin{vmatrix} 1 & 1 & 6 \\ \alpha & \beta & 3 \\ 1 & 2 & 14 \end{vmatrix} \\ D_z = 1(14\beta - 6) - 1(14\alpha - 3) + 6(2\alpha - \beta) \\ D_z = 14\beta - 6 - 14\alpha + 3 + 12\alpha - 6\beta \\ D_z = 8\beta - 2\alpha - 3 $$ Step 3: Determine the conditions for unique solution, no solution, and infinitely many solutions. Find the specific values of $(\alpha, \beta)$ for infinitely many solutions. For a system of linear equations: * **Unique Solution**: $D \neq 0$. This means $2\beta - \alpha - 7 \neq 0$. * **No Solution**: $D = 0$ and at least one of $D_x, D_y, D_z$ is non-zero. This means $2\beta - \alpha - 7 = 0$ AND ($4\beta + 11 \neq 0$ OR $-4\alpha - 50 \neq 0$ OR $8\beta - 2\alpha - 3 \neq 0$). * **Infinitely Many Solutions**: $D = 0$ AND $D_x = 0$ AND $D_y = 0$ AND $D_z = 0$. Let's find the values of $\alpha$ and $\beta$ that satisfy all these conditions: 1. $2\beta - \alpha - 7 = 0$ 2. $4\beta + 11 = 0 \implies \beta = -11/4$ 3. $-4\alpha - 50 = 0 \implies \alpha = -50/4 = -25/2$ Now, check if these values satisfy the remaining conditions: For $D=0$: $2(-11/4) - (-25/2) - 7 = -11/2 + 25/2 - 7 = 14/2 - 7 = 7 - 7 = 0$. (Satisfied) For $D_z=0$: $8(-11/4) - 2(-25/2) - 3 = -22 + 25 - 3 = 0$. (Satisfied) Thus, the system has infinitely many solutions if and only if $(\alpha, \beta) = (-25/2, -11/4)$. Step 4: Evaluate Option 1. **Option 1: If $\alpha=\beta=7$, the system has no solution.** Substitute $\alpha=7, \beta=7$ into $D$: $D = 2(7) - 7 - 7 = 14 - 14 = 0$. Since $D=0$, we check $D_x$: $D_x = 4\beta + 11 = 4(7) + 11 = 28 + 11 = 39$. Since $D=0$ and $D_x \neq 0$, the system has no solution. Therefore, Option 1 is TRUE. Step 5: Evaluate Option 2. **Option 2: If $\alpha=\beta$ and $\alpha\neq7$, the system has unique solution.** Substitute $\alpha=\beta$ into $D$: $D = 2\alpha - \alpha - 7 = \alpha - 7$. If $\alpha \neq 7$, then $D = \alpha - 7 \neq 0$. When $D \neq 0$, the system has a unique solution. Therefore, Option 2 is TRUE. Step 6: Evaluate Option 3. **Option 3: There is a unique point $(\alpha,\beta)$ on $x+2y+18=0$ for which the system has infinitely many solutions.** We found that the system has infinitely many solutions only for the unique point $(\alpha, \beta) = (-25/2, -11/4)$. Now, we check if this point lies on the line $x+2y+18=0$. Substitute $\alpha = -25/2$ for $x$ and $\beta = -11/4$ for $y$: $(-25/2) + 2(-11/4) + 18 = -25/2 - 11/2 + 18 = -36/2 + 18 = -18 + 18 = 0$. Since the point satisfies the equation of the line, it lies on the line. As this is the only point for infinitely many solutions, Option 3 is TRUE. Step 7: Evaluate Option 4 and state the conclusion. **Option 4: For every point $(\alpha,\beta)\neq(7,7)$ on $x-2y+7=0$, the system has infinitely many solutions.** The equation of the line $x-2y+7=0$ implies $\alpha - 2\beta + 7 = 0$, which can be rewritten as $\alpha = 2\beta + 7$. Substituting this into the expression for $D$: $D = 2\beta - \alpha - 7 = 2\beta - (2\beta + 7) - 7 = 2\beta - 2\beta - 7 - 7 = -14$. Wait, there is a mistake in my earlier deduction: $\alpha = 2\beta - 7$ was from $D=0$. Let's recheck $D=2\beta-\alpha-7$. If $D=0$, then $\alpha = 2\beta-7$. The line in Option 4 is $x-2y+7=0$, which is $\alpha-2\beta+7=0$. This is equivalent to $\alpha = 2\beta-7$. If $(\alpha, \beta)$ lies on this line, then $D = 2\beta - \alpha - 7 = 2\beta - (2\beta - 7) - 7 = 2\beta - 2\beta + 7 - 7 = 0$. So, any point $(\alpha, \beta)$ on the line $x-2y+7=0$ makes $D=0$. For the system to have infinitely many solutions, *in addition* to $D=0$, all of $D_x, D_y, D_z$ must also be zero. We found that this only occurs at the unique point $(\alpha, \beta) = (-25/2, -11/4)$. Let's check if this point lies on $x-2y+7=0$: $(-25/2) - 2(-11/4) + 7 = -25/2 + 11/2 + 7 = -14/2 + 7 = -7 + 7 = 0$. So the unique point for infinite solutions *is* on the line $x-2y+7=0$. However, the statement says that *every* point $(\alpha, \beta) \neq (7,7)$ on this line yields infinitely many solutions. This is incorrect. Consider a point on $x-2y+7=0$ different from $(-25/2, -11/4)$ and $(7,7)$. For example, if we choose $\beta = 0$, then $\alpha = 2(0) - 7 = -7$. So, consider the point $(\alpha, \beta) = (-7, 0)$. This point is on $x-2y+7=0$. For this point, $D=0$ (as calculated above). Now check $D_x = 4\beta + 11 = 4(0) + 11 = 11$. Since $D=0$ and $D_x \neq 0$, the system has no solution for $(\alpha, \beta) = (-7, 0)$. Therefore, it is not true that *every* point $(\alpha,\beta)\neq(7,7)$ on $x-2y+7=0$ results in infinitely many solutions. Option 4 is FALSE. The final answer is $\boxed{\text{Option 4}}$