Using the Jordan canonical form gives a second way to see this. Since the algebraic multiplicity of $\lambda$ is 3, the sizes of all Jordan blocks belonging to $\lambda$ add up to 3. Since the geometric multiplicity is 2, the number of Jordan blocks belonging to $\lambda$ is exactly 2 (the number of blocks always equals the geometric multiplicity).
Two block sizes that are positive integers and add up to 3 must be $1$ and $2$ (three blocks of size 1 would instead give geometric multiplicity 3). So the Jordan structure for $\lambda$ consists of one $1\times1$ block and one $2\times2$ block.
A matrix is diagonalizable exactly when every Jordan block has size 1, that is, the Jordan form itself is a diagonal matrix. Since a $2\times2$ block is present here, the Jordan form of $M$ is not diagonal, so $M$ is not similar to any diagonal matrix.
\[\boxed{M \text{ is not diagonalizable}}\]