System of equation x + 3y + 2z = 6 , $x+\lambda y+2z=7$ , $\mathrm{x}+3\mathrm{y}+2\mathrm{z}=\mu \text{ has }$

$\begin{array}{l}x+3y+2z=6----\left(i\right)\\ x+\lambda y+2z=7----\left(ii\right)\\ x+3y+2z=\mu ----\left(iii\right)\end{array}$

$\text{ A) If }\lambda =2\text{, then }D=0,\text{ therefore unique solution is not possible }$

$\begin{array}{l}\text{ B) If }\lambda =4,\mu =6\\ x+3y=6-2z\\ x+4y=7-2z\\ y=1\text{ and }x=3-2z\end{array}$

Substituting in equation (iii) 
3 – 2z + 3 + 2z = 6 is satisfied 
  Infinite solutions
$\text{ C) }\lambda =5,\mu =7$

$\begin{array}{l}\text{ Consider equation (ii) and (iii) }\\ x+5y=7-2z\end{array}$

$\begin{array}{l}x+3y=7-2z\\ \therefore y=0x=7-2z\text{ are solution }\\ \text{ Sub. In (i) }\end{array}$

$7-2z+2z=6\text{ does not satisfy }$

$\begin{array}{l}\text{ no solution }\\ \text{ D) if }\lambda =3,\mu =5\end{array}$

then equation (i) and (ii) have no solution 

no solution