Given five planes:
$\pi_1=2x+5y+z-2=0$
$\pi_2=x+y-z-1=0$
$\pi_3=x+4y+2z-4=0$
$\pi_4=3x-y+4z-3=0$
$\pi_5=-6x+2y-8z+k=0$.
How can i find the solid shape that is formed by those planes?
I tried to draw but it's too complex.
I can see that for a solid to be exists $k\neq6$.
Any help?
Thanks.
How $\pi_2$ , $\pi_3$ are independents planes and $\pi_1$ in linear combination of $\pi_2$ and $\pi_3$ but the except to independent term so is a prisma triagular infinity.
How the plane $\pi_4$ , $\pi_5$ and $k\neq{6}$ are parallel planes then is a triangular prism bounded for this planes.