Question:

Step 1: Understand the condition for infinitely many solutions. For a system of linear equations of the form $A\mathbf{x} = \mathbf{b}$ to have infinitely many solutions, two conditions must generally be met: 1. The determinant of the coefficient matrix, $|A|$, must be zero. 2. If $|A|=0$, then for infinitely many solutions, the determinants $D_x, D_y, D_z$ (formed by replacing a column of the coefficient matrix with the constant terms) must also be zero. Step 2: Form the coefficient matrix and calculate its determinant. The given system of equations is: $x+2y+3z=3$ $4x+3y-4z=4$ $8x+4y-\lambda z=9+\mu$ The coefficient matrix $A$ is: $$ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{pmatrix} $$ Now, we calculate the determinant of $A$: $$ |A| = 1 \begin{vmatrix} 3 & -4 \\ 4 & -\lambda \end{vmatrix} - 2 \begin{vmatrix} 4 & -4 \\ 8 & -\lambda \end{vmatrix} + 3 \begin{vmatrix} 4 & 3 \\ 8 & 4 \end{vmatrix} $$ $$ |A| = 1(3(-\lambda) - (-4)(4)) - 2(4(-\lambda) - (-4)(8)) + 3(4(4) - 3(8)) $$ $$ |A| = 1(-3\lambda + 16) - 2(-4\lambda + 32) + 3(16 - 24) $$ $$ |A| = -3\lambda + 16 + 8\lambda - 64 + 3(-8) $$ $$ |A| = 5\lambda - 48 - 24 $$ $$ |A| = 5\lambda - 72 $$ Step 3: Set the determinant of the coefficient matrix to zero to find $\lambda$. For infinitely many solutions, $|A|$ must be zero: $$ 5\lambda - 72 = 0 $$ $$ 5\lambda = 72 $$ $$ \lambda = \frac{72}{5} $$ Step 4: Form the determinant $D_z$ and calculate it. For infinitely many solutions, we must also have $D_x=0$, $D_y=0$, and $D_z=0$. Let's use $D_z$, which is the determinant of the matrix formed by replacing the third column of $A$ with the constant terms vector $\begin{pmatrix} 3 \\ 4 \\ 9+\mu \end{pmatrix}$: $$ D_z = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 3 & 4 \\ 8 & 4 & 9+\mu \end{vmatrix} $$ Now, we calculate $D_z$: $$ D_z = 1 \begin{vmatrix} 3 & 4 \\ 4 & 9+\mu \end{vmatrix} - 2 \begin{vmatrix} 4 & 4 \\ 8 & 9+\mu \end{vmatrix} + 3 \begin{vmatrix} 4 & 3 \\ 8 & 4 \end{vmatrix} $$ $$ D_z = 1(3(9+\mu) - 4(4)) - 2(4(9+\mu) - 4(8)) + 3(4(4) - 3(8)) $$ $$ D_z = 1(27+3\mu - 16) - 2(36+4\mu - 32) + 3(16 - 24) $$ $$ D_z = (11+3\mu) - 2(4+4\mu) + 3(-8) $$ $$ D_z = 11+3\mu - 8 - 8\mu - 24 $$ $$ D_z = -5\mu - 21 $$ Step 5: Set $D_z$ to zero to find $\mu$. For infinitely many solutions, $D_z$ must be zero: $$ -5\mu - 21 = 0 $$ $$ -5\mu = 21 $$ $$ \mu = -\frac{21}{5} $$ Step 6: State the final ordered pair. Based on our calculations, the values for $\lambda$ and $\mu$ are $\frac{72}{5}$ and $-\frac{21}{5}$ respectively. Thus, the ordered pair $(\lambda, \mu)$ is $\left(\frac{72}{5}, -\frac{21}{5}\right)$. The final answer is $\left(\dfrac{72}{5},\dfrac{-21}{5}\right)$. The correct option is Option 3.