In a rectangular cuboid $ABCDA_1B_1C_1D_1$ the face diagonals are $7, 8, 9$. What is the height from vertex $B$ of the $ABCB_1$ pyramid?

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The vertex $B$ is surrounded by $A, C, B_1$

I'm stuck here. I feel like the problem can be solved with analytical geometry by searching from the distance between a point and a plane, but that hasn't been part of our program so I assume it can be done much more simply.

I can calculate the sides of the cuboid from the diagonals, but all my attempts after that have resulted in the wrong solution. I feel like the projection of $B$ on the $B1AC$ plane is an important point on a triangle, but I'm not sure if it has to be and if it is, which one.

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Let $h$ be our height, $BA=x$, $BB_1=y$ and $BC=z$.

Hence, $S_{\Delta ACB_1}=\sqrt{12\cdot5\cdot4\cdot3}=12\sqrt5$.

In another hand, $x^2+y^2=49$, $x^2+z^2=64$ and $y^2+z^2=81$, which gives

$x^2+y^2+z^2=97$ and $x=4$, $y=\sqrt{33}$ and $z=4\sqrt3$.

Thus, $\frac{1}{3}h\cdot12\sqrt5=\frac{1}{6}\cdot4\cdot\sqrt{33}\cdot4\sqrt3$, which gives the answer: $$h=2\sqrt{\frac{11}{5}}$$