In fact, for any surface $\Sigma$, there exists a matrix $A_0 \in SL_3(\mathbb{R})$ so that $A_0\cdot\Sigma$ is a critical point for variations by $sl(3)$. The point is that if we look at the orbit $\{A\cdot\Sigma, A\in SL_3(\mathbb{R})\}$, the area function $\Phi: SL_3(\mathbb{R})\to \mathbb{R}$ given by $\Phi(A)=Area(A\cdot\Sigma)$ is proper on this orbit. That is, as $A \to \infty$, $\Phi(A)\to \infty$. Granted this, then there is an absolute minium of the function $\Phi$ at some matrix $A_0\in SL_3(\mathbb{R})$. Hence $A_0\cdot \Sigma$ will be a critical point for $\Phi(A(t))$ for any path $A(t)\in SL_3(\mathbb{R})$, $A(0)=A_0$.

To see the claim, first note that it is true for $\Sigma_r=S^2_r$, a round sphere of radius $r$. The orbit $A\cdot S^2_r$ consists of ellipsoids of axes $a,b, c $ such that $abc=r^3$. Then the area of these ellipsoids $\to \infty$ if (say) $a \to \infty$, by an approximate area formula for ellipsoids.

For the general case, assume that $\Sigma$ is an embedded closed surface containing the sphere $S^2_r$ bounding the ball $B_r$. For any $A\in SL_3(\mathbb{R})$, consider the nearest point projection $\rho_{r,A}: \mathbb{R}^3 \to A\cdot B_r$. Then this map is area decreasing, since the derivative is a linear projection at each point in $\mathbb{R}^3-A\cdot B_r$. Hence $Area(A\cdot \Sigma) \geq Area(\rho_{r,A} A\cdot \Sigma) \geq Area(A\cdot S^2_r)$. So the area function $\Phi$ is proper since it is for ellipsoids.