The values of $$\lambda$$ and $$\mu$$ for which the system of linear equations $$x + y + z = 2$$, $$x + 2y + 3z = 5$$, $$x + 3y + \lambda z = \mu$$ has infinitely many solutions, are respectively:

We are given the three equations

$$\begin{aligned} x+y+z &= 2,\\ x+2y+3z &= 5,\\ x+3y+\lambda z &= \mu. \end{aligned}$$

For a system of three linear equations in three unknowns to possess infinitely many solutions, two conditions must hold:

1. The coefficient matrix must be singular, that is, its determinant must vanish. (Rank < 3.)

2. The rank of the augmented matrix must be the same as the rank of the coefficient matrix, otherwise the system would be inconsistent.

We first write the coefficient matrix $$A$$ and compute its determinant.

$$A=\begin{bmatrix} 1&1&1\\ 1&2&3\\ 1&3&\lambda \end{bmatrix}.$$

The determinant of a $$3\times3$$ matrix

$$\begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix}$$

is given by the expansion

$$\det A = a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31}).$$

Applying this formula to our matrix we have

$$\begin{aligned} \det A &= 1\bigl(2\lambda-3\cdot3\bigr)\;-\;1\bigl(1\cdot\lambda-3\cdot1\bigr)\;+\;1\bigl(1\cdot3-2\cdot1\bigr)\\ &= 1\bigl(2\lambda-9\bigr)\;-\;\bigl(\lambda-3\bigr)\;+\;\bigl(3-2\bigr)\\ &= 2\lambda-9-\lambda+3+1\\ &= \lambda-5. \end{aligned}$$

For the determinant to vanish we set

$$\lambda-5 = 0 \quad\Longrightarrow\quad \lambda = 5.$$

Thus the first condition imposes $$\lambda=5.$$

Now we check the augmented matrix to ensure consistency. Substituting $$\lambda=5$$ into the original system we get

$$\begin{aligned} x+y+z &= 2,\\ x+2y+3z &= 5,\\ x+3y+5z &= \mu. \end{aligned}$$

We perform elementary row operations. First subtract the first equation from the second and third to eliminate $$x$$:

$$\begin{aligned} \text{(Row2)} &\;-\;\text{(Row1)}:\; & (x+2y+3z) - (x+y+z) &= y+2z &= 5-2 &= 3,\\[4pt] \text{(Row3)} &\;-\;\text{(Row1)}:\; & (x+3y+5z) - (x+y+z) &= 2y+4z &= \mu-2. \end{aligned}$$

So, after these operations the simplified system is

$$\begin{aligned} x+y+z &= 2,\\ y+2z &= 3,\\ 2y+4z &= \mu-2. \end{aligned}$$

Notice that the third new equation is simply twice the second new equation provided that the right‐hand sides scale in the same way. Indeed, multiplying the second new equation by $$2$$ gives

$$2(y+2z) = 2\cdot3 \;\;\Longrightarrow\;\; 2y+4z = 6.$$

For the third new equation to match this exactly we require

$$\mu-2 = 6 \quad\Longrightarrow\quad \mu = 8.$$

Thus consistency forces $$\mu=8.$$

Both conditions are now satisfied: the determinant is zero (making the coefficient matrix singular) and the augmented matrix has the same rank (two), giving infinitely many solutions.

Therefore, the required values are $$\lambda = 5,\;\; \mu = 8.$$ Hence, the correct answer is Option C.