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

We are given the three equations

$$x + y + z = 5$$

$$x + 2y + 2z = 6$$

$$x + 3y + \lambda z = \mu.$$

For a system of three linear equations in the three variables $$x,y,z$$ to possess infinitely many solutions, the three equations must be consistent and linearly dependent, which means the third equation must be expressible as a linear combination of the first two. Equivalently, every solution of the first two equations must automatically satisfy the third one.

Start by eliminating $$x$$ using the first equation. From

$$x + y + z = 5$$

we have

$$x = 5 - y - z.$$

Next, subtract the first equation from the second:

$$\bigl(x + 2y + 2z\bigr) - \bigl(x + y + z\bigr) = 6 - 5,$$

which simplifies term-by-term to

$$y + z = 1.$$

This relation can be rewritten as

$$z = 1 - y.$$

Now substitute $$x = 5 - y - z$$ into the third equation:

$$x + 3y + \lambda z = \mu$$

$$\Downarrow$$

$$(5 - y - z) + 3y + \lambda z = \mu.$$

Combine like terms carefully:

$$5 - y - z + 3y + \lambda z = 5 + 2y + (\lambda - 1)z.$$

Because every solution of the first two equations satisfies $$z = 1 - y$$, we substitute $$z = 1 - y$$ into this expression:

$$5 + 2y + (\lambda - 1)(1 - y) = \mu.$$

Expand the last term:

$$5 + 2y + (\lambda - 1) - (\lambda - 1)y = \mu.$$

Group constant and $$y$$ terms:

$$\bigl[5 + (\lambda - 1)\bigr] \;+\; y\bigl[2 - (\lambda - 1)\bigr] = \mu.$$

Simplify the coefficient of $$y$$:

$$2 - (\lambda - 1) = 2 - \lambda + 1 = 3 - \lambda.$$

Hence the left side is

$$5 + (\lambda - 1) + y(3 - \lambda).$$

For the expression to be independent of the free parameter $$y$$—and thus for the third equation to hold for every solution of the first two—the coefficient of $$y$$ must vanish. Therefore, we must impose

$$3 - \lambda = 0,$$

which yields

$$\lambda = 3.$$

With $$\lambda = 3$$, the entire left-hand side becomes purely constant:

$$5 + (3 - 1) = 5 + 2 = 7.$$

Thus we must have

$$\mu = 7.$$

Finally, compute the required sum:

$$\lambda + \mu = 3 + 7 = 10.$$

Hence, the correct answer is Option B.