In order for assumption (2) to hold, the elements $v_1, v_2, v_3$ must be $k$-linearly independent in $V$ (small detail omitted). Thus we may choose a basis $v_1, \ldots, v_r$ of $V$ incorporating these elements as the first $3$. Recall that $H_{univ} \subset \mathbf{A}^r_k \times_k X$ is the ``universal divisor''. Consider the projection $q : H_{univ} \to \mathbf{A}^r_k$ whose scheme theoretic fibres are the divisors $H_v$. By Lemma \ref{lemma-geom-irreducible-generic-fibre} it suffices to show that the generic fibre of $q$ is geometrically irreducible. To prove this we may replace $X$ by its reduction, hence we may assume $X$ is an integral scheme of finite type over $k$. \medskip\noindent Let $U \subset X$ be a nonempty affine open such that $\mathcal{L}|_U \cong \mathcal{O}_U$. Write $U = \Spec(A)$. Denote $f_i \in A$ the element corresponding to section $\psi(v_i)|_U$ via the isomorphism $\mathcal{L}|_U \cong \mathcal{O}_U$. Then $H_{univ} \cap (\mathbf{A}^r_k \times_k U)$ is given by $$ H_U = \Spec(A[x_1, \ldots, x_r]/(x_1f_1 + \ldots + x_rf_r)) $$ By our choice of basis we see that $f_1$ cannot be zero because this would mean $v_1 = 0$ and hence $H_{v_1} = X$ which contradicts assumption (2). Hence $\sum x_if_i$ is a nonzerodivisor in $A[x_1, \ldots, x_r]$. It follows that every irreducible component of $H_U$ has dimension $d + r - 1$ where $d = \dim(X) = \dim(A)$. If $U' = U \cap D(f_1)$ then we see that $$ H_{U'} = \Spec(A_{f_1}[x_1, \ldots, x_r]/(x_1f_1 + \ldots + x_rf_r)) \cong \Spec(A_{f_1}[x_2, \ldots x_r]) = \mathbf{A}^{r - 1}_k \times_k U' $$ is irreducible. On the other hand, we have $$ H_U \setminus H_{U'} = \Spec(A/(f_1)[x_1, \ldots, x_r]/(x_2f_2 + \ldots + x_rf_r)) $$ which has dimension at most $d + r - 2$. Namely, for $i \not = 1$ the scheme $(H_U \setminus H_{U'}) \times_U D(f_i)$ is either empty (if $f_i = 0$) or by the same argument as above isomorphic to an $r - 1$ dimensional affine space over an open of $\Spec(A/(f_1))$ and hence has dimension at most $d + r - 2$. On the other hand, $(H_U \setminus H_{U'}) \times_U V(f_2, \ldots, f_r)$ is an $r$ dimensional affine space over $\Spec(A/(f_1, \ldots, f_r))$ and hence assumption (3) tells us this has dimension at most $d + r - 2$. We conclude that $H_U$ is irreducible for every $U$ as above. It follows that $H_{univ}$ is irreducible. \medskip\noindent Thus it suffices to show that the generic point of $H_{univ}$ is geometrically irreducible over the generic point of $\mathbf{A}^r_k$, see Varieties, Lemma \ref{varieties-lemma-geometrically-irreducible-function-field}. Choose a nonempty affine open $U = \Spec(A)$ of $X$ contained in $X \setminus H_{v_1}$ which meets the irreducible component $Z$ of $H_{v_2} \cap H_{v_3}$ whose existence is averted in assumption (2). With notation as above we have to prove that the field extension $$ \text{Frac}(A[x_1, \ldots, x_r]/(x_1f_1 + \ldots + x_rf_r))/ k(x_1, \ldots , x_r) $$ is geometrically irreducible. Observe that $f_1$ is invertible in $A$ by our choice of $U$. Set $K = \text{Frac}(A)$ equal to the fraction field of $A$. Eliminating the variable $x_1$ as above, we find that we have to show that the field extension $$ K(x_2, \ldots, x_r)/ k(x_2, \ldots, x_r, -\sum\nolimits_{i = 2, \ldots, r} f_1^{-1}f_i x_i) $$ is geometrically irreducible. By Lemma \ref{lemma-amazing-bertini-lemma} it suffices to show that for some $c_2, \ldots, c_r \in k$ the elements $$ x_2, \ldots, x_r, \sum\nolimits_{i = 2, \ldots, r} f_1^{-1}f_i x_i, \sum\nolimits_{i = 2, \ldots, r} c_if_1^{-1}f_i $$ are algebraically independent over $k$ in the fraction field of $A[x_2, \ldots, x_r]$. This follows from Lemma \ref{lemma-algebraically-independent} and the fact that $Z \cap U$ is an irreducible component of $V(f_1^{-1}f_2, f_1^{-1}f_3) \subset U$.