If the system of linear equations,

Solution:

$x+y+z=6$.....(i)

$x+2 y+3 z=10$...(ii)

$3 x+2 y+\lambda z=\mu$...(iii)

From (i) and (ii),

If $z=0 \Rightarrow x+y=6$ and $x+2 y=10$

$\Rightarrow y=4, x=2$

$(2,4,0)$

If $y=0 \Rightarrow x+z=6$ and $x+3 z=10$

$\Rightarrow \quad z=2$ and $x=4$

$(4,0,2)$

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

$(4,0,2)$

So, $6+8=\mu \Rightarrow \mu=14$

and $12+2 \lambda=\mu$

$12+2 \lambda=14 \Rightarrow \lambda=1$

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