Using the exact sequence

$$0\mapsto\mathcal{O}_{\mathbb{P}^{2}}\rightarrow\mathcal{O}_{\mathbb{P}^{2}}(1)^{\oplus 3}\rightarrow T_{\mathbb{P}^{2}}\mapsto 0$$

it is easy to compute $H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = H^{2}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = 0$ while $h^{0}(\mathbb{P}^{2},T_{\mathbb{P}^{2}}) = 8$.

On the singular variety $\mathbb{P}(1,2,3)$ by $T_{\mathbb{P}(1,2,3)}$ I mean $\mathcal{H}om(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})$.

**Is there an analogous way (or a completely different way) of computing the cohomology groups of $T_{\mathbb{P}(1,2,3)} = \mathcal{H}om(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})$ ?**

If it helps $\mathbb{P}(1,2,3)$ can be embedded in $\mathbb{P}^{6}$ as a singular Del Pezzo surface of degree six.