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.

$x+y+z=6 ...\left(1\right)$
$x+2y+3z=10 ...\left(2\right)$
$3x+2y+\lambda z=\mu \dots \left(3\right)$

From $\left(1\right) \& \left(2\right)$,
If $z=0\Rightarrow x+y=6$ and $x+2y=10$, $\Rightarrow y=4, x=2$
If $y=0\Rightarrow x+z=6$ and $x+3z=10$, $\Rightarrow z=2, x=4$

So, $3x+2y+\lambda z=\mu$, must pass through $(\mathrm{2,4},0)$ and $(\mathrm{4,0},2)$.

$6+8=\mu \Rightarrow \mu =14$
And $12+2\lambda =\mu \Rightarrow \lambda =1$

So, $\mu -{\lambda }^2=14-1$$=13$