We have the equations
\begin{eqnarray*}
x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 1\\
x_1 x_2 + x_2 x_3 + x_4 x_5 + x_5 x_6 = 0 \\
x_1 x_3 + x_4 x_6 = 0
\end{eqnarray*}
multipy the second and third equations by $2$ and add them to the first equation gives $(x_1+x_2+x_3)^2+(x_4+x_5+x_6)^2=1$ which can be parameterised to
\begin{eqnarray*}
x_1+x_2+x_3=\cos \theta \\
x_4+x_5+x_6 = \sin \theta . \\
\end{eqnarray*}
Now substitute these into the second equation and we have
\begin{eqnarray*}
x_2 (\cos \theta-x_2)+ x_5( \sin \theta-x_5)=0 , \\
\end{eqnarray*}
multiply this equation by $-4$ and complete the squares,
\begin{eqnarray*}
(2x_2 -\cos \theta)^2+ (2x_5- \sin \theta)^2=1 , \\
\end{eqnarray*}
which can be parameterised to give
\begin{eqnarray*}
2x_2=\cos \theta +\cos \phi \\
2(x_1+x_3)=\cos \theta -\cos \phi \\
2x_5 = \sin \theta+\sin \phi \\
2(x_4+x_6) = \sin \theta-\sin \phi . \\
\end{eqnarray*}
Now substitute these into the third equation and we have
\begin{eqnarray*}
x_1 \left(\frac{\cos \theta-\cos \phi}{2}-x_1\right)+ x_4\left(\frac{ \sin \theta-\sin \phi}{2}-x_4\right)=0 , \\
\end{eqnarray*}
multiply this equation by $-16$ and complete the squares,
\begin{eqnarray*}
(4x_1 -(\cos \theta-\cos \phi))^2+ (4 x_4-( \sin \theta-\sin \phi)^2=(\cos \theta-\cos \phi))^2+ ( \sin \theta-\sin \phi)^2 =2(1- \cos(\theta- \phi)). \\
\end{eqnarray*}
Thus the manifold can be fully parameterised by
\begin{eqnarray*}
4x_1 &=& \cos \theta -\cos \phi+\sqrt{2(1- \cos(\theta- \phi)} \cos \psi \\
4x_3 &=& \cos \theta -\cos \phi+\sqrt{2(1- \cos(\theta- \phi)} \cos \psi \\
4x_4 &=& \sin \theta -\sin \phi+\sqrt{2(1- \cos(\theta- \phi)} \sin \psi \\
4x_6 &=& \sin \theta -\sin \phi+\sqrt{2(1- \cos(\theta- \phi)} \sin \psi \\
2x_2 &=& \cos \theta +\cos \phi \\
2x_5 &=& \sin \theta+\sin \phi. \\
\end{eqnarray*}
None of these equations have any discontiunities and so to obtain a path from $(\theta, \phi, \psi)$ to $(\theta', \phi', \psi')$ the simple parameterisation will do $(t\theta +(1-t)\theta', t\phi+(1-t)\phi', t\psi +(1-t) \psi')$ for $t \in [0,1]$. So the space is path connected, As far characterisation is concerned it is $3-$dimensional apart from when $ \theta= \phi$ when the space is $1-$dimensional(forming a unit circle in the $x_2-x_5$ plane.)