If the system of linear equations x + y + z = 5 , x + 2y + 2z = 6 ,...

∆ = \left|\begin{cases}1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda\end{cases}\right| = \left(\lambda - 3\right)
{∆}_{1} = \left|\begin{cases}5 & 1 & 1 \\ 6 & 2 & 2 \\ \mu & 3 & \lambda\end{cases}\right| = 4\left(\lambda - 3\right)
{∆}_{3} = \left|\begin{cases}1 & 5 & 1 \\ 1 & 6 & 2 \\ 1 & \mu & \lambda\end{cases}\right| = \lambda - \mu + 4
{∆}_{4} = \left|\begin{cases}1 & 1 & 5 \\ 1 & 2 & 6 \\ 1 & 3 & \mu\end{cases}\right| = \mu - 7
For infinite many solution $∆ = {∆}_{1} = {∆}_{2} = {∆}_{3} = 0$
$⇒ \lambda = 3, \text{ and } \mu = 7$ from $∆ = 0 \text{ and } {∆}_{4} = 0$
& $\lambda = 3, \mu = 7$ Satisfies ${∆}_{3} = 0 = {∆}_{1}$
$∴ \lambda + \mu = 10$