Question:

Step 1: Identify the condition for infinitely many solutions. For a system of linear equations $AX=B$ with $n$ variables, there are infinitely many solutions if and only if $\text{det}(A)=0$ and $\text{det}(A_i)=0$ for all $i=1,2,\dots,n$, where $A_i$ is the matrix formed by replacing the $i$-th column of $A$ with the column matrix $B$. This is often described as $\text{rank}(A) = \text{rank}([A|B]) < n$. Step 2: Form the coefficient matrix and calculate its determinant. The given system of equations is: $$ x-y+z=5 $$ $$ 2x+2y+\alpha z=8 $$ $$ 3x-y+4z=\beta $$ The coefficient matrix $A$ is: $$ A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{pmatrix} $$ Calculate the determinant of $A$: $$ \text{det}(A) = 1(2 \cdot 4 - \alpha \cdot (-1)) - (-1)(2 \cdot 4 - \alpha \cdot 3) + 1(2 \cdot (-1) - 2 \cdot 3) $$ $$ \text{det}(A) = 1(8+\alpha) + 1(8-3\alpha) + 1(-2-6) $$ $$ \text{det}(A) = 8+\alpha + 8-3\alpha - 8 $$ $$ \text{det}(A) = 8-2\alpha $$ Step 3: Use the condition $\text{det}(A)=0$ to find $\alpha$. For infinitely many solutions, $\text{det}(A)$ must be equal to zero: $$ 8-2\alpha = 0 $$ $$ 2\alpha = 8 $$ $$ \alpha = 4 $$ Step 4: Form the determinant $D_x$ (or $D_1$) by replacing the first column of $A$ with the constant terms. To ensure consistency and infinitely many solutions (since $\text{det}(A)=0$), we must have $D_x=0$. The matrix for $D_x$ is formed by replacing the first column of $A$ with the constant terms from the right-hand side of the equations: $$ D_x = \begin{vmatrix} 5 & -1 & 1 \\ 8 & 2 & \alpha \\ \beta & -1 & 4 \end{vmatrix} $$ Substitute the value $\alpha=4$ into $D_x$: $$ D_x = \begin{vmatrix} 5 & -1 & 1 \\ 8 & 2 & 4 \\ \beta & -1 & 4 \end{vmatrix} $$ Calculate the determinant $D_x$: $$ D_x = 5(2 \cdot 4 - 4 \cdot (-1)) - (-1)(8 \cdot 4 - 4 \cdot \beta) + 1(8 \cdot (-1) - 2 \cdot \beta) $$ $$ D_x = 5(8+4) + 1(32-4\beta) + 1(-8-2\beta) $$ $$ D_x = 5(12) + 32-4\beta - 8-2\beta $$ $$ D_x = 60 + 24 - 6\beta $$ $$ D_x = 84 - 6\beta $$ Step 5: Use the condition $D_x=0$ to find $\beta$. For infinitely many solutions, $D_x$ must be equal to zero: $$ 84 - 6\beta = 0 $$ $$ 6\beta = 84 $$ $$ \beta = 14 $$ Step 6: Form the quadratic equation whose roots are $\alpha$ and $\beta$. We found $\alpha=4$ and $\beta=14$. A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1+r_2)x + r_1r_2 = 0$. Sum of the roots: $\alpha + \beta = 4 + 14 = 18$. Product of the roots: $\alpha \beta = 4 \times 14 = 56$. Therefore, the quadratic equation is: $$ x^2 - 18x + 56 = 0 $$ Step 7: Conclude the final answer. The values of $\alpha$ and $\beta$ are $4$ and $14$ respectively. These are the roots of the equation $x^2-18x+56=0$. Comparing this with the given options, it matches Option 3. The final answer is $\boxed{x^2-18x+56=0}$.