PDE $u_{x} ^{2} +u_{y}^{2}+u_{z}^{2}=1$ with constant data on plane 
$$u_{x} ^{2} +u_{y}^{2}+u_{z}^{2}=1, \qquad  u=k \;\text{ on the plane }\; \alpha x+\beta y + z =0, \text{ } $$
  where $k$, $\alpha$ and $\beta$ are constants

I've been trying to solve this PDE problem but I can only get to solve for $u_{x}, u_{y}$, and $u_{z}$ in terms of $\alpha$ and $\beta$. Can anyone help me out? Thanks!
 A: Inspection works, but you may solve this more systematically by generalising Charpit's method to more than two independent variables. Parametrise everything using $s_1, s_2, t$ where the initial data is given by:
$(x(s_1,s_2,0),y(s_1,s_2,0),z(s_1,s_2,0)) = (s_1,s_2,-\alpha s_1 - \beta s_2)$. With the dot denoting differentiation by $t$, writing $F(p,q,r)=p^2+q^2+r^2-1 \equiv 0$ we have:
\begin{equation}
\dot{x}= F_p = 2p, \\
\dot{y} = F_q =2q, \\
\dot{z} = F_r =2r, \\
\dot{p} = \dot{q} = \dot{r} = 0, \\
\dot{u} = pF_p + qF_q + rF_r = 2(p^2+q^2+r^2) = 2.
\end{equation}
From the penultimate line, $p(s_1,s_2,t) = p_0(s_1,s_2)$ and analogously for $q$ and $r$. But the initial data now gives us three equations that determine $p_0,q_0, r_0$: one is from the PDE itself at the data curve, and two come from differentiating along it with respect to each of the two parameters. Namely:
\begin{equation}
p_{0}^{2} + q_{0}^{2} + r_{0}^{2} = 1, \\
\dfrac{\partial u_0}{\partial s_1} = \dfrac{\partial u_0}{\partial x_0}\dfrac{\partial x_0}{\partial s_1} + \dfrac{\partial u_0}{\partial y_0}\dfrac{\partial y_0}{\partial s_1} + \dfrac{\partial u_0}{\partial z_0}\dfrac{\partial z_0}{\partial s_1} = p_0 - \alpha r_0, \\
\dfrac{\partial u_0}{\partial s_2} = \dfrac{\partial u_0}{\partial x_0}\dfrac{\partial x_0}{\partial s_2} + \dfrac{\partial u_0}{\partial y_0}\dfrac{\partial y_0}{\partial s_2} + \dfrac{\partial u_0}{\partial z_0}\dfrac{\partial z_0}{\partial s_2} = q_0 - \beta r_0.
\end{equation}
Therefore, since $u$ is just a constant initially, $p_0 = \alpha r_0$, $q_0 = \beta r_0$, so from the sum of squares
\begin{equation}
r_0 = \pm \dfrac{1}{\sqrt{1+ \alpha^2 + \beta^2}}.
\end{equation}
Now everything may be recovered into a parametric solution:
\begin{equation}
x = \pm \dfrac{2\alpha t}{\sqrt{1+ \alpha^2 + \beta^2}} + s_1, \\
y = \pm \dfrac{2\beta t}{\sqrt{1+ \alpha^2 + \beta^2}} + s_2, \\
z = \pm \dfrac{2 t}{\sqrt{1+ \alpha^2 + \beta^2}}-\alpha s_1 - \beta s_2, \\
u = 2 t + k.
\end{equation}
Now notice that:
\begin{equation}
\dfrac{\alpha x + \beta y + z}{\sqrt{1+ \alpha^2 + \beta^2}} = \pm 2 t = \pm (u-k).
\end{equation}
This gives us two solutions:
\begin{equation}
u(x,y,z) = \dfrac{\alpha x + \beta y + z}{\sqrt{1+ \alpha^2 + \beta^2}} + k,
\end{equation}
or the alternative
\begin{equation}
u(x,y,z) = -\dfrac{\alpha x + \beta y + z}{\sqrt{1+ \alpha^2 + \beta^2}} + k.
\end{equation}
