Question:

Step 1: Write down the given system of linear equations and form the coefficient matrix. The given system of linear equations is: $$x + y + z = 6 \quad \text{(1)}$$ $$x + 2y + 3z = 10 \quad \text{(2)}$$ $$x + 2y + \lambda z = \mu \quad \text{(3)}$$ The coefficient matrix $A$ for this system is: $$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{pmatrix}$$ Step 2: State the condition for a system of linear equations to have infinitely many solutions. For a system of linear equations $AX = B$ to have infinitely many solutions, the determinant of the coefficient matrix, $\det(A)$, must be zero. Also, the determinants of the matrices formed by replacing a column of $A$ with the constant terms (known as $D_x, D_y, D_z$ or $\det(A_x), \det(A_y), \det(A_z)$) must also be zero. For a $3 \times 3$ system, a simpler approach when $\det(A) = 0$ is to check if one equation becomes a linear combination of the others or is identical, leading to dependent equations. Step 3: Calculate the determinant of the coefficient matrix $A$. We calculate the determinant of matrix $A$: $$\det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix}$$ Expanding along the first row: $$\det(A) = 1 \cdot (2\lambda - 3 \cdot 2) - 1 \cdot (1 \cdot \lambda - 3 \cdot 1) + 1 \cdot (1 \cdot 2 - 2 \cdot 1)$$ $$\det(A) = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2)$$ $$\det(A) = 2\lambda - 6 - \lambda + 3 + 0$$ $$\det(A) = \lambda - 3$$ Step 4: Apply the condition $\det(A) = 0$ to find the value of $\lambda$. For infinitely many solutions, we must have $\det(A) = 0$. $$\lambda - 3 = 0$$ $$\lambda = 3$$ Step 5: Substitute the value of $\lambda$ back into the system and determine the value of $\mu$ for infinitely many solutions. Substitute $\lambda = 3$ into the third equation of the system: $$x + 2y + 3z = \mu \quad \text{(3')}$$ Now, compare equation (3') with equation (2): Equation (2): $x + 2y + 3z = 10$ Equation (3'): $x + 2y + 3z = \mu$ For the system to have infinitely many solutions when $\det(A) = 0$, the equations must be consistent and dependent. In this case, if the third equation is identical to the second equation, they will represent the same plane, leading to infinitely many solutions (a line of intersection with the first plane). Therefore, for equation (3') to be consistent and yield infinitely many solutions with equation (2), their right-hand sides must be equal: $$\mu = 10$$ Step 6: Conclude the values of $\lambda$ and $\mu$. For the system of linear equations to have infinitely many solutions, we must have $\lambda = 3$ and $\mu = 10$. The final answer is $\lambda = 3, \mu = 10$. The correct option is Option 1.