The set of all values of λ for which the system of linear...

$\left(2 - \lambda\right) {x}_{1} - 2 {x}_{2} + {x}_{3} = 0$

$2 {x}_{1} - \left(\lambda + 3\right) {x}_{2} + 2 {x}_{3} = 0$

$- {x}_{1} + 2 {x}_{2} - \lambda {x}_{3} = 0$

The systems of linear equations will have a non-trivial solution

⇒ \left|\begin{cases}2 - \lambda & - 2 & 1 \\ 2 & - \lambda - 3 & 2 \\ - 1 & 2 & - \lambda\end{cases}\right| = 0

$⇒ \left(2 - \lambda\right) \left[{\lambda}^{2} + 3 \lambda - 4\right] + 2 \left[- 2 \lambda + 2\right] + 1 \left[4 - \lambda - 3\right] = 0$

$2 {\lambda}^{2} + 6 \lambda - 8 - {\lambda}^{3} - 3 {\lambda}^{2} + 4 \lambda - 4 \lambda + 4 + 1 - \lambda = 0$

$- {\lambda}^{3} - {\lambda}^{2} + 5 \lambda - 3 = 0$

${\lambda}^{3} + {\lambda}^{2} - 5 \lambda + 3 = 0$

\left(\lambda - 1\right) \left({\lambda}^{2} + 2 \lambda - 3\right) = 0$<br /><br />\left(\lambda - 1\right) \left(\lambda + 3\right) \left(\lambda - 1\right) = 0$

$⇒ \lambda = -3 , 1 , 1$