Is this a diagonalizable matrix?

The characteristic polynomial is $$\chi_A(x)=\det(A-xI)$$ and its roots are the eigenvalues of the matrix $A$. By the hypothesis we see that $0,-2$ and $3$ are the $3$ distinct eigenvalues so $A$ is diagonalizable and since $A$ has $0$ as eigenvalue with multiplicity $1$ then $A$ isn't invertible and $$\dim\ker A=1$$ so by the rank-nullity theorem the rank of $A$ is $2$.

The three determinants that are zero tell you that $0$, $-2$ and $+3$ are roots of the characteristic polynomial. Since your matrix is $3\times3$, the degree of that chacteristic polynomial is$~3$, which forces all those roots to be simple roots, and the characteristic polynomial to be $X(X+2)(X-3)$. In the case of simple roots (and only in that case) you can know without further computation what the dimension of the corresponding eigenspace will be, namely$~1$, so here you have eigenspaces for $0$, $-2$ and $+3$, each of dimension$~1$. Together they span the whole $3$-dimensional space, so $A$ is diagonalizable. The $1$-dimensional eigenspace for $\lambda=0$ is the kernel of $A$, which matrix is therefore is not invertible and has rank$~3-1=2$.

Note that all conclusions depend on having as much distinct eigenvalues as the dimension of the space; without that (for instance if only two of those zero determinants were given) nothing could have been said.

By definition, $\lambda$ is an eigenvalue if there exists some nonzero $x$ such that $Ax=\lambda x$.

This is equivalent to the existance of a nonzero $x$ such that $(A-\lambda I) x = 0$, which is equivalent to $\det(A-\lambda I) = 0$.

Conclusion: if $\det(A-\lambda I) = 0$, then $\lambda$ is an eigenvalue of $A$.

Plugging $\lambda = 0$ into the conclusion also answers your final question.