Why are vinylic and arylic carbocations highly unstable?
Background

S-orbitals are lower in energy than p-orbitals, therefore the more s-character in an orbital, the lower its energy and the more stable any electrons occupying the orbital.  In other words, electrons will be more stable (lower in energy) in an $\ce{sp}$ orbital than an $\ce{sp^2}$ orbital, and electrons in an $\ce{sp^2}$ orbital will be lower in energy than those in an $\ce{sp^3}$ orbital.

This is the basis of the inductive effect.  For example, it explains why the electrons in an $\ce{sp^3}$ methyl group will be drawn towards a connected $\ce{sp^2}$ carbon in an aromatic ring - the methyl group is inductively electron donating to the aromatic ring as its $\ce{sp^3}$ electrons drift towards the lower energy $\ce{sp^2}$ aromatic carbon.

What we're really saying is that an $\ce{sp}$ hybridized carbon atom is more electronegative (it's lower in energy and the electrons prefer to move towards it) than an $\ce{sp^2}$ hybridized carbon atom, which in turn is more electronegative than an $\ce{sp^3}$ hybridized carbon atom.

Answer

Look at the figure below, notice that in the alkyl carbocation on the left the cationic center is attached to an $\ce{sp^3}$ carbon, whereas in the vinylic cation in the middle, the cationic center is attached to a more electronegative $\ce{sp^2}$ carbon.  For a positive center to be attached to a more electronegative group is destabilizing.  Hence the vinylic cation is less stable than a typical alkyl cation.

Things are even worse with the aryl carbocation on the right.  Here the positive carbon is attached to 2 $\ce{sp^2}$ carbons. Destabilizing the aryl cation even further is its geometry.  A vinyl cation prefers to be linear, but due to geometrical constraints imposed by the aromatic ring the aryl cation must be bent and the empty orbital is forced to be $\ce{sp^2}$ rather than $\ce{p}$.  These three factors combine to make the aryl carbocation even higher in energy than the vinyl cation.