\begin{align*} & \left| \begin{array}{ccc} 1-\lambda & -1 & 2\\ 2 & -2-\lambda & 4\\ 3 & -3 & 6-\lambda\end{array}\right|=0\\ \implies & \left| \begin{array}{ccc} -\lambda & -1 & 0\\ \lambda & -2-\lambda & -\lambda\\ 0& -3 & -\lambda\end{array}\right|=0 \qquad \text{Applying } C1=C1+C2, \& \; C3=C3+2C2\\ \implies & \lambda=0,0,2 \end{align*} But $\lambda=2$ does not satisfies this equation.
If we expand determinant directly we get $\lambda=0,0,5$ and these satisfies the equation.
Anyone tell me where is my mistake?
Updated:. \begin{align*} & \left| \begin{array}{ccc} 1-\lambda & -1 & 2\\ 2 & -2-\lambda & 4\\ 3 & -3 & 6-\lambda\end{array}\right|=0\\ \implies & \left| \begin{array}{ccc} -\lambda & -1 & 0\\- \lambda & -2-\lambda & -2\lambda\\ 0& -3 & -\lambda\end{array}\right|=0 \qquad \text{Applying } C1=C1+C2, \& \; C3=C3+2C2\\ \implies & \lambda^2\left| \begin{array}{ccc} 1 & 1 & 0\\ 1 & 2+\lambda & 2\\ 0& 3 & 1\end{array}\right|=0\\ \implies & \lambda=0,0,5 \end{align*}
In the entry $\;2,3\;$ after you applied those elementary operations on the columns, it must be $\;-2\lambda\;$ , and also in $\;2,1\;$ it must be $\;\color{red}{-\lambda}\;$