Consider the plane in R^4 , calculate an orthonormal basis 

Let's first rewrite the two equations to make them easier to handle. By replacing them with half of the sum and the difference, we get \[ x +2z + w = 0; \qquad y + 3z - 2w = 0 \] so to get a basis for $M$ we just pick any two values for $z$ and $w$ and compute the corresponding values of $x$ and $y$. For $v_1$, we can set e.g. $z=0$ and $w=1$, and get \[ u_1 = (-1, 2, 0, 1); \] for $v_2$, we can set e.g. $z=1$ and $w=0$, and get \[ u_2 = (-2, -3, 1, 0). \] Clearly $u_1$ and $u_2$ are independent and belong to $M$, so they form a basis of $M$. To get a basis of the complement, we can pick any two independent vectors such that the space they generate only intersect $M$ in the origin. We can take e.g. \[ u_3 = (1,0,0,0); \qquad u_4 = (0,1,0,0). \] These are clearly independent and any non-zero linear combination does not belong to $M$, as replacing $(x,y,0,0)$ in the equations of the plane yields $x=0$ and $y=0$.

Now to get an orthogonal basis we just apply the Gram-Schmidt algorithm (https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process) to the basis $\{u_1, u_2, u_3, u_4 \}$ and get the basis \[ v_1 = \frac{1}{\sqrt 6} (-1,2,0,1), \qquad v_2 = \frac{1}{\sqrt{102}} (-8, -5, 3, 2) \] and \[ w_1 = \frac{1}{\sqrt{238}} (7, -2, 8, 11), \qquad w_2 = \frac{1}{\sqrt{14}}(0,1,3,-2) \] (I had the computer do the calculation, but you can do that by hand - it's annoying though). A simple check ensures that they are indeed orthnormal, and that $v_1$ and $v_2$ belong to $M$ (and so they span it).