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\begin{center}
\begin{tikzpicture}[
    % Viewing angle for looking slightly up at the tetrahedron from the side
    x={(1cm,0.cm)},      % Horizontal view (side perspective)
    y={(0.2cm,0.1cm)},    % Very slight vertical compression (looking up)
    z={(0cm,1cm)},   % Strong z-projection to enhance "looking up" effect
    scale=3
]
    % Define the coordinates with B, C, D on a "table" (z=0) and A above
    \coordinate (B) at (0, 0, 0);       % Left front corner on table
    \coordinate (C) at (1, 0, 0);       % Right front corner on table
    \coordinate (D) at (0.5, 1, 0);     % Back corner on table
    \coordinate (A) at (0.5, 0.333, 0.8); % Above the center of the base
    
    % Calculate the center of mass (centroid)
    \coordinate (CM) at ($0.25*(A)+0.25*(B)+0.25*(C)+0.25*(D)$);
    
    % Draw the back edges with dashed lines
    \draw[thick] (A) -- (C);
    \draw[thick, dashed] (A) -- (D);
    \draw[thick, dashed] (C) -- (D);
    
    % Draw the front edges with solid lines
    \draw[thick] (A) -- (B) -- (C) -- cycle;
    \draw[thick, dashed] (B) -- (D);
    
    % Draw the lines connecting vertices to the center of mass
    \draw[red] (A) -- (CM);
    \draw[red] (B) -- (CM);
    \draw[red] (C) -- (CM);
    \draw[red] (D) -- (CM);
    
    % Label the vertices and center of mass
    \fill (A) circle (0.03) node[above] {$A$};
    \fill (B) circle (0.03) node[left] {$B$};
    \fill (C) circle (0.03) node[right] {$C$};
    \fill (D) circle (0.03) node[above right] {$D$};
    \fill[red] (CM) circle (0.6pt) node[right] {$X$};
    
    % Add a caption
    \node[below=1cm] at (current bounding box.south) {Tetrahedron with center of mass and all connecting lines};
\end{tikzpicture}
\end{center}

The distance of $A$ to $X$ is $\frac34$ the distance from $A$ to the centre base ($d$)

The distance of $C$ to the centre of the base ($G$) is $\frac{2}{3}$ the height of $BCD$ which is $\frac{\sqrt{3}}{2} \cdot 2a = \sqrt{3} a$.

Therefore we must have $(2a)^2 = d^2 + \frac43a^2 \Rightarrow d = \frac{2\sqrt{2}}{\sqrt{3}}a$ and so $AX = \frac34  \frac{2\sqrt{2}}{\sqrt{3}}a = \sqrt{\frac32}a$

The tension in each string will be $\lambda \left (\sqrt{\frac32}-1 \right)$.

Considering the energy of the system, when the ABCD reaches it's maximum height, it's velocity will be $0$. Therefore the only energies to consider are GPE and EPE.

Assuming the table is $0$, we initially have $EPE$ of

\begin{align*}
3 \cdot \frac12 \lambda \frac{(a(\sqrt{\frac32}-1))^2}{a} = \frac32 \lambda a\left (\frac52-2\sqrt{\frac32} \right )
\end{align*}

When $BCD$ is level with $O$, the height is $\frac{1}{\sqrt{6}}a$ and GPE of $\frac{Mga}{\sqrt{6}}$

The $EPE$ will be:

\begin{align*}
3 \cdot \frac12 \lambda \frac{(a(\frac{2}{\sqrt{3}}-1))^2}{a} &= \frac32 \lambda a \left (\frac73 - \frac{4}{\sqrt{3}}\right )
\end{align*}

So by conservation of energy:

\begin{align*}
&& \frac32 \lambda a\left (\frac52-2\sqrt{\frac32} \right ) &= \frac{Mga}{\sqrt{6}} + \frac32 \lambda a \left (\frac73 - \frac{4}{\sqrt{3}}\right ) \\
\Rightarrow && \frac{Mg}{\lambda} &= \sqrt{6} \left (\frac32 \left (\frac52-2\sqrt{\frac32} \right )  -  \frac32 \left (\frac73 - \frac{4}{\sqrt{3}}\right ) \right) \\
&&&= -9 + 6\sqrt{2}+\sqrt{\frac38} \\
&&&= 0.09765380\ldots \\
&&&= 0.098\, (2\text{ s.f})
\end{align*}

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