Investigate for what values of \(\lambda\) and \(\mu\) the system of linear equations \(x+y+z=6,\; x+2y+3z=10,\; x+2y+\lambda z=\mu\) will have (i) No solution, (ii) Unique solution, (iii) Infinite number of solutions.

Augmented matrix: \([A\mid B]=\left[\begin{array}{ccc|c}1&1&1&6\\1&2&3&10\\1&2&\lambda&\mu\end{array}\right]\)

\(R_2\to R_2-R_1,\; R_3\to R_3-R_1\): \(\left[\begin{array}{ccc|c}1&1&1&6\\0&1&2&4\\0&1&\lambda-1&\mu-6\end{array}\right]\)

\(R_3\to R_3-R_2\): \(\left[\begin{array}{ccc|c}1&1&1&6\\0&1&2&4\\0&0&\lambda-3&\mu-10\end{array}\right]\)

(i) No solution: when \(\rho(A)\ne\rho([A|B])\), i.e. \(\rho(A)=2,\ \rho([A|B])=3\): \(\lambda=3,\ \mu\ne10\).

(ii) Unique solution: when \(\rho(A)=\rho([A|B])=3\) (number of unknowns): \(\lambda\ne3\), \(\mu\) any value.

(iii) Infinitely many solutions: when \(\rho(A)=\rho([A|B])=2<3\): \(\lambda=3,\ \mu=10\). Then \(z=k\) (arbitrary), \(y=4-2k\), \(x=6-y-z=2+k\).