Problem source

The tetrahedron $ABCD$ has $A$ at the point $(0,4,-2)$. It is symmetrical about the plane $y+z=2,$ which passes through $A$ and $D$. The mid-point of $BC$ is $N$. The centre, $Y$, of the sphere $ABCD$ is at the point $(3,-2,4)$ and lies on $AN$ such that $\overrightarrow{AY}=3\overrightarrow{YN}.$ Show that $BN=6\sqrt{2}$ and find the coordinates of $B$ and $C$. 
The angle $AYD$ is $\cos^{-1}\frac{1}{3}.$ Find the coordinates of $D$. 
[There are two alternative answers for each point.]

Solution source

Since $B$ and $C$ are reflections of each other in the plane $y+z=2$ (since that's what it means to be symmetrical), we must have that $N$ also lies on the plane $y+z=2$.

Since $\overrightarrow{AY}=3\overrightarrow{YN}.$ we must have $\overrightarrow{AN}=\overrightarrow{AY}+\overrightarrow{YN} = \frac43\overrightarrow{AY} = \frac43\begin{pmatrix} 3\\-6\\6\end{pmatrix} = \begin{pmatrix} 4\\-8\\8\end{pmatrix}$ and $N$ is the point $(4,-4,6)$ (which fortunately is on our plane as expected).  $Y$ is the point $(3,-2,4)$

$|\overrightarrow{AY}| = \sqrt{3^2+(-6)^2+6^2} = 3\sqrt{1+4+4} = 9$

\begin{center}
    \begin{tikzpicture}
    \coordinate (Y) at (0,0);
    \coordinate (A) at (-3, 0);
    \coordinate (N) at ({1},0);
    \coordinate (B) at ({1}, {3*sin(acos(1/3))});

    \draw (Y) circle (3);
    \filldraw (Y) circle (1pt) node[below] {$Y$};
    \filldraw (A) circle (1pt) node[left] {$A$};
    \filldraw (N) circle (1pt) node[below] {$N$};
    \filldraw (B) circle (1pt) node[above] {$B$};

    \draw (A) -- (N) -- (B);
    \draw (Y) -- (B) -- (A);

    \node[below] at ($(A)!0.5!(Y)$) {$9$};
    \node[below] at ($(N)!0.5!(Y)$) {$3$};
    \node[left] at ($(B)!0.5!(Y)$) {$9$};
    
    
    \end{tikzpicture}
\end{center}

Notice that $BN^2 + 3^2 = 9^2 \Rightarrow BN^2 = 3\sqrt{3^2-1} = 6\sqrt2$.

Therefore $\overrightarrow{NB} = \pm 6\sqrt{2} \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}$ and $\{ B, C\} =\{ (4, 2, 12), (4, -10, 0)\}$.



Suppose $D = (x,y,z)$ then 

\begin{align*}
&& \begin{pmatrix} -1 \\ 2 \\ -2\end{pmatrix} \cdot \begin{pmatrix} x- 3 \\ y+2 \\ z-4\end{pmatrix} &= 3  \cdot 9 \cdot \frac13 = 9\\
\Rightarrow && 9 &= 3-x+2(y+2)-2(z-4) \\
&&&= -x+2y-2z+15 \\
\Rightarrow && 6 &= x-2y+2z \\
&& 2 &= x -4y \\
\\
\Rightarrow && 81 &= (4y+2-3)^2+(y+2)^2+(2-y-4)^2 \\
&&&= (4y-1)^2+2(y+2)^2 \\
&&&= 16y^2-8y+1+2y^2+8y+8 \\
&&&= 18y^2+9 \\
\Rightarrow && y^2 &= 2 \\
\Rightarrow && y &= \pm 2
\end{align*}

Therefore $\displaystyle D \in \left \{ (10, 2, 0), (-6, -2, 4) \right \}$