We may assume that $S_1 \ge S_2 \ge S_3$. We shall prove that the answer is $S_1$.
It is not difficult to see that the area of the convex hull of two segments $d_1$ and $d_2$ is not less than the area $S$ of a triangle with two sides equal and parallel to these segments. Indeed, if this hull is a triangle, two cases are possible. In the first case, the respective segments are sides and then the area is equal to $S$. In the second case, we may assume that an interior point $E$ of $\triangle ABC$ is such that $CE = d_1$ and $AB = d_2$. Setting $F = AB \cap CE$, then
$$2S_{ABC} = AB \cdot CF \sin \angle CFB \ge d_1 d_2 \sin \angle CFB = 2S.$$

If the hull is a quadrilateral with diagonals the respective segments, then its area is equal to $S$. It remains to consider the case when the hull is a quadrilateral with two opposite sides the respective segments. We may assume that points $C$ and $D$ on the sides $BE$ and $AE$ of $\triangle ABE$ are such that $AD = d_1$ and $BC = d_2$. Then
$$2S_{ABCD} = 2S_{ABE} - 2S_{CDE} = (AE \cdot BE - DE \cdot CE) \sin \angle AEB > AD \cdot BC \sin \angle AEB = 2S.$$
Hence any convex polygon containing two segments equal and parallel to $XB$ and $XC$, is not less than $S_1$.
Let now $A'$ be the symmetric point of $A$ with respect to $X$, and $D$ be such that $XBDC$ is a parallelogram. Then $A'$ lies either in $\triangle BXD$, or in $\triangle CXD$; for example, in $\triangle BXD$. Since $S_{XCA'} = S_2 \le S_1$, then $A'$ lies in $\triangle BXD$. This triangle has area $S_1$ and contains three segments equal and parallel to $XA$, $XB$, $XC$.