Distance from $A$ to $\triangle BCD$ if $\angle BAC=\angle CAD=\angle DAB=60^\circ$? Let $A$, $B$, $C$, and $D$ be four points in space such that
$$\angle BAC=\angle CAD=\angle DAB=60^\circ.$$
If $AB=1$, $AC=2$, and $AD=6$, then what is the distance between $A$ and the plane of $\triangle BCD$?

Observations: $ABC$ and $BCD$ are right, as $ABC$ is $30-60-90$ and then using Law of Cosines we han find the other side lengths for $BCD.$ I don't know how to use this to solve for the distance desired though. I have tried letting this distance be $h$ and setting up equations.

You can use the formula for tetrahedron volume (wiki),
$V = \frac{abc}{6} \sqrt{1+2 \cos \alpha \cos \beta \cos \gamma - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma}$
$\alpha, \beta, \gamma$ are angles between edges at a vertex and $a, b, c$ are lengths of edges from the vertex.
Here we know that at vertex $A$, $\angle BAC = \angle BAD = \angle CAD = 60^0$ and $AB = 1, AC = 2, AD = 6$.
So $V = \displaystyle \frac{1\cdot2\cdot6}{6} \sqrt{1 + \frac{2}{2^3} - \frac{3}{2^2}} = \sqrt2$
We also know that the volume of Tetrahedron is given by $V = \frac{1}{3} A \cdot h$ where $A$ is the area of the base and $h$ is the altitude.
As you found out, $\triangle BCD$ is right angled triangle with $BC = \sqrt3, CD = \sqrt{28}, BD = \sqrt{31}$. So, $A = \sqrt{21}$.
So altitude from $A$ to base BCD, $h = \frac{3V}{A} = \sqrt{\frac{6}{7}}$
