If the system of linear equations,
$$x + y + z = 6$$
$$x + 2y + 3z = 10$$
$$3x + 2y + \lambda z = \mu$$
has more than two solutions, then $$\mu - \lambda^2$$ is equal to

We have the system

$$x + y + z = 6$$

$$x + 2y + 3z = 10$$

$$3x + 2y + \lambda z = \mu$$

First we solve the first two equations to understand the set of common solutions they already permit.

From $$x + y + z = 6$$ and $$x + 2y + 3z = 10$$ we subtract the first from the second to eliminate $$x$$:

$$\bigl(x + 2y + 3z\bigr) - \bigl(x + y + z\bigr) = 10 - 6$$

$$y + 2z = 4$$

So we can express $$y$$ in terms of $$z$$:

$$y = 4 - 2z$$

Now we substitute this value of $$y$$ back into the first equation $$x + y + z = 6$$:

$$x + (4 - 2z) + z = 6$$

$$x + 4 - z = 6$$

$$x = 2 + z$$

Hence every solution of the first two equations can be written as

$$x = 2 + z, \qquad y = 4 - 2z, \qquad z \text{ is free}$$

Thus the first two equations already give infinitely many solutions (one free parameter $$z$$).

For the entire system to have “more than two solutions”, the third equation must not reduce the number of free parameters. Therefore every triple $$\bigl(2 + z,\; 4 - 2z,\; z\bigr)$$ must also satisfy the third equation. We now impose this condition.

The third equation is $$3x + 2y + \lambda z = \mu$$. Substituting $$x = 2 + z$$ and $$y = 4 - 2z$$ we obtain

$$3(2 + z) + 2(4 - 2z) + \lambda z = \mu$$

First expand each term:

$$3 \times 2 = 6,\;\; 3 \times z = 3z$$

$$2 \times 4 = 8,\;\; 2 \times (-2z) = -4z$$

So

$$6 + 3z + 8 - 4z + \lambda z = \mu$$

Combine the constants and the coefficients of $$z$$:

$$14 + (3z - 4z + \lambda z) = \mu$$

$$14 + (\lambda - 1)z = \mu$$

This equality must hold for all values of the free parameter $$z$$. The only way a linear expression in $$z$$ can equal a constant $$\mu$$ for every $$z$$ is if the coefficient of $$z$$ is zero. Therefore

$$(\lambda - 1) = 0 \quad\Longrightarrow\quad \lambda = 1$$

With $$\lambda = 1$$, the expression becomes simply

$$14 = \mu$$

So

$$\mu = 14$$

Now we compute the required quantity $$\mu - \lambda^2$$:

$$\mu - \lambda^2 = 14 - (1)^2 = 14 - 1 = 13$$

So, the answer is $$13$$.