Suppose, toward a contradiction, that we had a 2-CNF formula $\phi$, using the variables $x_1,x_2,x_3$, and some auxiliary variables $y_1,\dots,y_n$ such that, whenever we assign truth values to the $x$'s, if at least one of the three is true then we can assign values to the $y$'s to make $\phi$ true, but if all three of the $x$'s are false, then, no matter what values we give the $y$'s, $\phi$ will be false.
Choose an assignment $v_1$ of truth values to the $y$'s that makes $\phi$ true when $x_1$ is true but $x_2$ and $x_3$ are false. Similarly, choose assignments $v_2$ (resp. $v_3$) to the $y$'s that make $\phi$ true when only $x_2$ (resp. $x_3$) is true and the other two $x$'s are false. Let $v$ be the assignment of truth values to the $y$'s obtained by the majority vote of $v_1,v_2,v_3$, i.e., $v$ makes $y_i$ true iff at least two of $v_1,v_2,v_3$ make $y_i$ true.
By our assumption, if we give all three $x$'s the value false and give the $y$'s values according to $v$, then $\phi$ is false. So, as $\phi$ is a 2-CNF formula, one of the conjuncts in $\phi$ is false in this situation, and that conjunct is a disjunction of two literals (each of which is an $x$ or a $y$ or a negation of one of these.) Each of those two literals has, in our current truth assignment ($v$ for the $y$'s and false for the $x$'s), the same truth value as in at least two of the three earlier truth assignments ($v_i$ for the $y$'s and only $x_i$ true among the $x$'s). So at least one of those three earlier assignments agrees with the current one on both of the literals in the conjunct under consideration. Thus, at least one of the three earlier truth assignments falsified that conjunct and therefore falsified $\phi$. That contradicts how we chose those earlier assignments.