If ker ${\cal M}^{T} $ is one-dimensional, there are two possibilities:
Case 1: $(\mathbf{u},\mathbf{v},M\mathbf{u})$ are linearly dependent; then $M\mathbf{v} $ is linearly independent of the other three, and we can construct the vector $\mathbf{w} $ orthogonal to all four vectors via $\mathbf{w}_{i} = \epsilon_{ijkl} \mathbf{u}_{j} \mathbf{v}_{k} (M\mathbf{v} )_{l} $.
Case 2: $(\mathbf{u},\mathbf{v},M\mathbf{u})$ are linearly independent; then $M\mathbf{v} $ is linearly dependent on the other three, and we can construct the vector $\mathbf{w} $ orthogonal to all four vectors via $\mathbf{w}_{i} = \epsilon_{ijkl} \mathbf{u}_{j} \mathbf{v}_{k} (M\mathbf{u} )_{l} $, and we can also add arbitrary multiples of $\epsilon_{ijkl} \mathbf{u}_{j} \mathbf{v}_{k} (M\mathbf{v} )_{l} $, as long as the result doesn't cancel.
We can furthermore consolidate the two cases into the expression $$ \mathbf{w}_{i} = \epsilon_{ijkl} \mathbf{u}_{j} \mathbf{v}_{k} [(M\mathbf{u} )_{l} + (M\mathbf{v} )_{l} +\mbox{sgn}(\epsilon_{mnop} \mathbf{u}_{n} \mathbf{v}_{o} (M\mathbf{u} )_{p} \epsilon_{mqrs} \mathbf{u}_{q} \mathbf{v}_{r} (M\mathbf{v} )_{s})(M\mathbf{v} )_{l}] $$ because in either case, it reduces to the form specific to that case already given. Here, the last term in the square brackets is a kludge to take into account the loophole pointed out by Willie Wong in comments, that the $M\mathbf{u} $ and $M\mathbf{v} $ components orthogonal to both $\mathbf{u} $ and $\mathbf{v} $ may cancel; this last term eliminates such a cancellation. Maybe one can make this less ugly and completely symmetric in $\mathbf{u} $, $\mathbf{v} $ again.