I'm not exactly sure how the different points matter. I believe $p=[1,1,-1]^T, [2,-1,0]^T, [3,0,-1]^T, [0,3,-2]^T$ are all feasible directions.
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Hint Feasible directions are a set of vectors along which you can move from some point $x$ staying feasible. Let's do one by example: take $x_a = (0,0,2)^T$ and notice that $x_a$ satisfies the equality constraint and the inequality constraints, so it is itself feasible. To keep feasible, you cannot decrease $x_1$ or $x_2$ components (lest the inequality constraints become violated) and you must stay on the plane $x_1 + 2x_2 + 3x_3 = 6$, so you cannot increase those components, without decreasing the third one, and the third one cannot increase by itself either. So whichever form your feasible directions take, they must look like $(a,b,c)$ with $c<0$ and $a,b>0$. Can you be more specific (they gotta satisfy a certain equality, you should be able to derive a 2D basis for them).
gt6989b
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@Laura Note $(2,1)$ satisfies first 2 constraints with equality. So let new point be $(2+a,1+b)$ then the second constraint implies $$(2+a)+(1+b) \le 3 \iff a + b \le 0.$$ Similarly, the second constraint implies $$5 \ge (2+a)^2+(1+b)^2 = 5 + 4a + a^2 + 2b + b^2.$$ Can you finish simplifying this, and solve system of 2 inequalities? – gt6989b Nov 23 '20 at 19:52
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@Laura https://math.stackexchange.com/questions/565448/feasible-direction-for-a-point – gt6989b Nov 23 '20 at 21:10