<s>[INST] Feasible point of a system of linear inequalities Let $P$ denote $(x,y,z)\in \mathbb R^3$, which satisfies the inequalities:
$$-2x+y+z\leq 4$$ $$x \geq 1$$ $$y\geq2$$ $$ z \geq 3 $$ $$x-2y+z \leq 1$$ $$ 2x+2y-z \leq 5$$
How do I find an interior point in $P$? 
Is there a specific method, or should I just try some random combinations and then logically find an interior point?
 [/INST] Label the planes and their intersection points as follows
$\begin{array}{llllllll} 
&\Pi_1:&x=1       & A & B & C &   &   & W & X\\ 
&\Pi_2:&y=2       & A & B &   & D &   & W &   & Y\\ 
&\Pi_3:&z=3       & A &   & C & D &   &   & X & Y & Z\\ 
&\Pi_4:&-2x+y+z=4 &   & B & C &   & E &   &   & Y & Z\\ 
&\Pi_5:&x-2y+z=1  &   & B &   & D & E &   & X &   & Z\\ 
&\Pi_6:&2x+2y-z=5 &   &   & C & D & E & W\\
\end{array}$
where $A=(1,2,3),B=(1,2,4),C=(1,3,3),D=(2,2,3),E=(2,3,5)$ are points which also satisfy all the inequalities and $W=(1,2,1),X=(1,\frac{3}{2},3),Y=(\frac{1}{2},2,3),Z=(0,1,3)$ fail to satisfy all the inequalities.
The five points $A,B,C,D,E$ are the vertices of a triangular bipyramid. The two pyramids are $ABCD$ and $BCDE$ with common base $BCD$ which is internal and not a face. The six faces are the planes $\Pi_i$.
The graphic below has the $z$-axis vertical, with $E$ the apex at the top ($z=5$), $B$ the vertex halfway down ($z=4$) and $\Pi_3=ACD$ the triangular base (at $z=3$). It has two hidden triangular faces - the base and the back $\Pi_6=CDE$. The light blue vertical face is $\Pi_1=ABC$, and the purple vertical face is $\Pi_2=ABD$. The other two visible faces are $\Pi_4,\Pi_5$.
The centroid of the polyhedron is $(1.4,2.4,3.6)$. It obviously satisfies the three inequalities related to $\Pi_1,\Pi_2,\Pi_3$. For $\Pi_4$ we have $-2.8+2.4+3.6=3.2<4$; for $\Pi_5$ we have $1.4-4.8+3.6=0.2<1$; and for $\Pi_6$ we have $2.8+4.8-3.6=4<5$.

</s>