Let us assume we want to find local extremal of a function of two variables $f(x,y)$. Also, let us assume that we have three equality constraints. This means that the Jacobian matrix, which is $2 \times 3$ can not have rank $3$ as the number of constraints. So the NDCQ conditions are never satisfied and it does not make any sense to compute the Lagrange system. How do we proceed here?
My attempt: If we have $3$ constraints I think we have two cases: either the constraint set is the null set because the constraint are nit compatible between them (their intersection is the empty set) or we have two linearly independent constraints, that means that we can reduce the lagrange problem to a problem where we have only two constraints, for example. Am I correct? Do you have any example?
There are 8 possibilities. Suppose the constraints are g(x,y)-0, h(x,y)=0, u(x,y)=0. Then $$ \nabla g=[0,0] or \nabla f = \lambda \nabla g $$ and $$ \nabla h=[0,0] or \nabla f = \alpha \nabla h $$ and $$ \nabla u=[0,0] or \nabla f = \beta \nabla u $$ At each of these 8 possibilities, all 3 constraints must be satisfied.