If the system of linear equations :
$$x + y + z = 6$$,
$$x + 2y + 5z = 10$$,
$$2x + 3y + \lambda z = \mu$$
has infinitely many solutions, then the value of $$\lambda + \mu$$ equals :

For a system of linear equations to have infinitely many solutions, the equations must be linearly dependent. In simple terms, one equation can be formed by a combination of the others.

1. Observe the Equations

Let's look at the given system:

1. $$x + y + z = 6$$
2. $$x + 2y + 5z = 10$$
3. $$2x + 3y + \lambda z = \mu$$
4. The coefficient of $$x$$ in Eq(3) is 2, which is the sum of coefficients of $$x$$ in Eq(1) and Eq(2) ($$1 + 1 = 2$$).
5. The coefficient of $$y$$ in Eq(3) is 3, which is the sum of coefficients of $$y$$ in Eq(1) and Eq(2) ($$1 + 2 = 3$$).
6. Comparing the $$z$$ coefficients: $$\lambda = 6$$
7. Comparing the constants: $$\mu = 16$$

2. Find the Relationship (The "Shortcut")

Notice the coefficients of $$x$$ and $$y$$ in the third equation:

For the system to have infinitely many solutions, the entire third equation must be the sum of the first two equations:

$$\text{Eq}(1) + \text{Eq}(2) = \text{Eq}(3)$$

3. Solve for $$\lambda$$ and $$\mu$$

By adding Eq(1) and Eq(2):

$$(x + x) + (y + 2y) + (1z + 5z) = (6 + 10)$$

$$2x + 3y + 6z = 16$$

Now, compare this result to the given Eq(3): $$2x + 3y + \lambda z = \mu$$

4. Final Calculation

The question asks for the value of $$\lambda + \mu$$:

$$\lambda + \mu = 6 + 16 = \mathbf{22}$$

Correct Option: C