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(C) For a system of linear equations to have infinitely many solutions,the third equation must be a linear combination of the first two equations. Let

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$10$

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(C) For a system of linear equations to have infinitely many solutions,the third equation must be a linear combination of the first two equations. Let the third equation be $L_3 = a L_1 + b L_2$. $x + 3y + \lambda z = \mu = a(x + y + z - 5) + b(x + 2y + 2z - 6)$. Comparing the coefficients of $x, y, z$ and the constant term: $a + b = 1$ (coefficient of $x$) $a + 2b = 3$ (coefficient of $y$) $a + 2b = \lambda$ (coefficient of $z$) $5a + 6b = \mu$ (constant term) Solving the first two equations: Subtracting $(a + b = 1)$ from $(a + 2b = 3)$ gives $b = 2$. Substituting $b = 2$ into $a + b = 1$ gives $a = -1$. Now,find $\lambda$ and $\mu$: $\lambda = a + 2b = -1 + 2(2) = 3$. $\mu = 5a + 6b = 5(-1) + 6(2) = -5 + 12 = 7$. Therefore,$\lambda + \mu = 3 + 7 = 10$.

If the system of linear equations $x + y + z = 5$,$x + 2y + 2z = 6$,and $x + 3y + \lambda z = \mu$ (where $\lambda, \mu \in \mathbb{R}$) has infinitely many solutions,then the value of $\lambda + \mu$ is:

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