Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:
Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....
Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $v\in V$ such that $\{v, Tv,..T^{n-1}v\}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.
Now conversely, if minimal and charteristic polynomial are same, then we have a minimal polynomial which is of degree $n$. Take $v\neq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider $\{v, Tv, T^2v,.. T^{n-1}v\}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.
Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of $\{(1,-1,3), (2,-2,-3)\}$