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

Put z = 0 we have
x + y = 6, x + 2y = 10 and 3x + 2y = $\mathrm{\mu }$
On solving
y = 4, x = 2 
$\therefore \mathrm{\mu }=6+8=14$
Put y = 0 we have
x + z = 6, x + 3z =10 and 3x + $\mathrm{\lambda }$z = $\mathrm{\mu }$
On solving
$\begin{array}{l}\mathrm{z}=2,\mathrm{x}=4\\ \therefore 12+2\mathrm{\lambda }=14\\ 2\mathrm{\lambda }=2\\ \mathrm{\lambda }=1\\ \therefore \mathrm{\mu }-{\mathrm{\lambda }}^2=14-1=13\end{array}$