I am looking for an example of a convex function $f:\mathbb{R}^2 \to \mathbb{R}$ whose graph is not below any intersecting plane.
More precisely, suppose $(\bar 0,0),v=(\bar v,f(\bar v)),w=(\bar w,f(\bar w))$ are three points on the graph of $f$.
Then the set of points inside the corresponding triangle are
$$ \{ tv+sw \, | \, t,s \in [0,1]\}=\{ \big(t\bar v+s\bar w,tf(\bar v)+sf(\bar w)\big) \, | \, t,s \in [0,1]\}$$
I am looking for a convex function which does not satisfy
$$ f(t\bar v+s\bar w) \le tf(\bar v)+sf(\bar w) \tag{1}$$ for all $t,s \in [0,1]$.
Note that if $t+s \le 1$ then $(1)$ holds, since then, by convexity
$$ f(t\bar v+s\bar w)=f(t\bar v+s\bar w+(1-t-s)\bar 0) \le tf(\bar v)+sf(\bar w) +(1-t-s)f(\bar 0)= tf(\bar v)+sf(\bar w).$$
Any strictly convex function will do. Consider what happens when $w$ approaches $v$: then $(1)$ becomes $f((t+s)v) \le (t+s)f(v)$, hence $f(\lambda v)\le \lambda f(v)$ for $0\le \lambda\le 2$. This can only be true if $f$ is linear on the segment from $0$ to $2v$.
In particular, $f(v)=|v|^2$ fails (1), for example, with $v=(1,0)$ and $w=(1,1)$.
But if you wanted (1) to fail for all $v,w$, that cannot happen. For example, if $w=-v$ and $t+s>1$, then $tu+sw = t'u+s'w$ where $t'=t-\delta$, $s'=s-\delta$, and $\delta = (t+s-1)/2$. Since $t'+s' = 1$, we have $$ f(t'u+s'w)\le t'f(u)+s'f(w) \le tf(u)+sf(w) $$ where the last inequality follows from $f(u)+f(w)\ge 0$.
A more geometric argument is that the failure of (1) for all $v,w$ means there is no parallelogram with three vertices on the boundary of a set $U = \{x:f(x)\le \ell(v)\}$, for any function $v$. This can't happen: whatever the convex set $U$ is, it contains a parallelogram with three vertices on the boundary.