3D points challenge

$P(a, b, 0)$ is the point that cuts the segment $AB$ two-thirds of the way from $A$ to $B$ . Let's construct the vector $\vec{AB} = \langle x + 3, y - 5, z - 8 \rangle$ . Two-thirds of this vector added to $A$ will give us $P$ from which we can easily deduce $z$ :

$\frac{2}{3}\langle x + 3, y - 5, z - 8 \rangle + (-3, 5, 8) = (a, b, 0) \longrightarrow z = -4$

In a similar fashion, $Q(0, 0, c)$ is the point for which $A$ is twice the vector $\vec{QB}$ added to $Q$ . We have that $\vec{QB} = \langle x, y, z - c \rangle$ . From this we can easily deduce $x$ and $y$ :

$2\langle x, y, z - c \rangle + (0, 0, c) = (-3, 5, 8) \longrightarrow x = -1.5, y = 2.5$

Therefore, $x + y + z = -1.5 + 2.5 - 4 = -3$ .

$\vec{AP}_z=-8\implies\vec{AB}_z=\frac32·(-8)=-12\implies B(x,y,8-12)=(x,y,-4)$ .

$B$ is the middle of the segment from $Q(0,0,?)$ to $A(-3,5,8)$ , so its coordinates are $B(\frac{0+(-3)}2,\frac{0+5}2,-4)=(\frac{-3}2,\frac52,-4)$ .