Let $G=Gr(2,6)$ the Grassmannian of two planes in $V=\mathbb C^6$, and let $\mathcal Q(1)$ the rank four quotient bundle on it twisted with $\mathcal O_G(1) \cong $ det$(S^*)$, $S$ being the tautological bundle.
Since $\mathcal Q(1)$ is globally generated, by adjunction the zero locus of a general global section $\lambda \in H^0(G, \mathcal Q(1))$ will be a smooth Fano fourfold of index 1.
As in this paper by Manivel, by Borel-Weil theorem, the space of global section $H^0(G, \mathcal Q(1))$ will admit a concrete description as $$H^0(G, \mathcal Q(1))= Ker \ (V \otimes \bigwedge^2 V^* \to V^*)$$ with the latter morphism being the (natural) contraction operator.
As said, by a theorem of Mukai, for a general $\lambda$, the variety $Z(\lambda$) will be smooth. The question now is
Q1: Given a specific $\lambda$, how can I check the smoothness of $Z(\lambda)$? (or the generality of $\lambda$ as element of $V \otimes \bigwedge^2 V^*$).
The given $\lambda$ I have to work with is in particular (with respect to the basis $v_1,\ldots, v_6$ for $\mathbb C^6$) $$\lambda= v_1 \otimes (v_2^* \wedge v_6^*+v_3^* \wedge v_5^*)+v_2 \otimes (v_3^* \wedge v_6^*+v_4^* \wedge v_5^*)+v_3 \otimes (v_1^* \wedge v_2^*+v_4^* \wedge v_6^*)++v_4 \otimes (v_1^* \wedge v_3^*+v_5^* \wedge v_6^*)+v_5 \otimes (v_1^* \wedge v_4^*+v_2^* \wedge v_3^*)+v_6 \otimes (v_1^* \wedge v_5^*+v_2^* \wedge v_4^*)$$
I have been able to check via computer algebra that $Z(\lambda)$ has indeed dimension 4, but smoothness is all another business. Moreover as an element of $V \otimes \bigwedge^2 V^* \cong Hom(\bigwedge^2 V,V)$, $\lambda$ can be represented by a matrix (with respect to the prescribed and induced bases) whose rank is maximal. Therefore the question Q1 might be replaced by
Q2: If $\lambda \in Hom\bigwedge^2 V,V) $ is represented by a matrix of maximal rank, can we conclude that $\lambda$ itself is "general"?