Ignoring the redundant fifth constraint, and plotting the feasible region in the $x$-$y$ plane, shows that it's bounded by a quadrilateral whose sides are the segments of the lines $\ -3y+4x=5\ $ between the points $\ (2,1)\ $ and $\ (5,5)\ $, $\ 4y-5x=15\ $ between the points $\ (5,5)\ $ and $\ (9,6)\ $, $\ y-3x=-21\ $ between the points $\ (9,6)\ $ and $\ (7,0)\ $, and $\ 5y+x=7\ $ between the points $\ (7,0)\ $ and $\ (2,1)\ $ (see the diagram below).
- For objectives that would have multiple extrema on this feasible set, you could take that of minimising $\ z=y-3x\ $, as you note in a comment, minimising $\ z=5y+x\ $, minimising $\ z=-3y+4x\ $, or maximising $\ z=4y-x\ $.
- Since the feasible region is bounded, there is no linear function which could be unbounded on it.
- I don't see how to make any sense of the third question. The feasible region doesn't depend in any way on the choice of objective, and since this particular feasible region is non-empty, no choice of objective is going to give a linear program that has no feasible solutions.