1. Write the system in matrix form:

For the system to have more than two solutions, it must have infinitely many solutions. That happens when the third equation is dependent on the first two.

2. First solve the first two equations partially.

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

$y+2z=4$

So, $y=4-2z$

Using $x+y+z=6$, $x=6-y-z=6-(4-2z)-z=2+z$

Thus the solutions of the first two equations are:

$x=2+z, \quad y=4-2z$

with $z$ free. Hence for infinitely many solutions, the third equation must be satisfied for all such $z$.

3. Substitute into the third equation:

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

Substitute $x=2+z$ and $y=4-2z$:

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

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

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

For this to hold for all values of $z$, the coefficient of $z$ must be zero:

$\lambda-1=0 \implies \lambda=1$

Then $\mu=14$

4. Now compute:

The expression in the question is $\mu-\lambda^2$.

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

5. Comparison with stored answer:

Stored correct answer = $13$, which matches our result.