Why a Rotation matrix $R$ don't "destroy" the Lorentz invariant
Prove what Mr. Avantgarde wrote
\begin{align*}
& \text{General Lorentz Transformation matrix}\\\\
\Lambda&=\begin{bmatrix}
\gamma & \gamma \,\vec{v}^T \\
\gamma \,\vec{v}^T & I+\frac{\gamma-1}{\vec{v}^2}\,\vec{v}\,\vec{v}^T \\
\end{bmatrix}\\\\
&\text{with:}\\
\gamma&=\frac{1}{\sqrt{1-\frac{\vec{v}^2}{c^2}}}\\\\
\vec{v}&=\begin{bmatrix}
v_x \\
v_y \\
v_z \\
\end{bmatrix}\,,\quad\text{velocity boost vector}\\
I&=\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix}\,,\quad\text{$3\times3$ identity matrix}\\\\
&\text{Line}\\
s^2&=x_\mu\,x^\mu=\eta_{\mu\nu}x^\nu\,x^\mu\\
s^2&=\vec{x}^T\,\eta\,\vec{x}=(x^0)^2-(x^1)^2-(x^2)-(x^3)\,,\quad \text{with}\\
\vec{x}&=\begin{bmatrix}
x^0 \\
\vec{x}^i \\
\end{bmatrix}\,,\quad \text{and}\\
\eta&=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}
\end{align*}
I) Lorentz transformation: Pure boost
\begin{align*}
&\text{line}\\
s'^2&=x'_\mu\,x'^\mu\,,\quad\text{with}\quad x'^\mu=\Lambda\,\vec{x}\quad\text{and}\quad x'_\mu=\eta\,\vec{x}\,\quad\Rightarrow \\
s'^2&=\vec{x}^T\,\underbrace{\Lambda^T\,\eta\,\Lambda}_{=\eta}\vec{x}= \vec{x}^T\,\eta\,\vec{x}=s^2\,,\quad\text{Lorentz invariant}
\end{align*}
II) Lorentz transformation: Boost plus rotation matrix $R$
\begin{align*}
\vec{x}&\mapsto
\begin{bmatrix}
x^0 \\
R\,\vec{x}^i \\
\end{bmatrix}\\
\quad \Rightarrow\\
&\text{line}\\
s'^2&=\vec{x}^T\,\underbrace{\Lambda^T\,\eta\,\Lambda}_{=\eta},\vec{x}= \vec{x}^T\,\eta\,\vec{x}=
\begin{bmatrix}
x^0 & \left(R\,\vec{x}^i\right)^T \\
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
x^0 \\
R\,\vec{x}^i \\
\end{bmatrix}\\
&=\begin{bmatrix}
x^0 & \vec{x^i}^T R^T \\
\end{bmatrix}
\begin{bmatrix}
x^0 \\
-R\,\vec{x}^i \\
\end{bmatrix}=(x^0)^2-(x^1)^2-(x^2)-(x^3)\,, \quad\text{again Lorentz invariant}\\
&\text{with}\quad R^T\,R=I
\end{align*}