How to determine whether a curve lies on a plane? 
Given that a sphere $x^2+y^2+z^2=1$ and a cylinder $x^2+y^2=x$ intersect at point $(1/2,1/2,1/\sqrt 2)$ determine whether the curve from the intersection lies on a plane.

We can find the curve using the following parametrization:
$$
c(t)=\bigg( \frac{1}{2}+\frac{1}{2}\cos t,\frac{1}{2}\sin t,\sin\frac{t}{2}\bigg)
$$
One of the ways to solve this is to make sure that tangent vectors to the curve are orthogonal to the plane $ax+by+cz=d$.
The tangent vector is:
$$
c'(t)=\bigg( -\frac{1}{2}\sin t,\frac{1}{2}\cos t,\frac{1}{2}\cos\frac{t}{2}\bigg)
$$
For $t=\pi$ we have $c'(t)=( 0,1/2,0)$.
Then:
$$
(a,b,c).(0,1/2,0)=0\implies b=0
$$
Similarly for $t=0\implies c=0$ and for $t=\pi/2\implies a=0$. Therefore such a plane doesn't exist.
I don't understand the logic of finding random points and why the fact that we found $a,b,c=0$ proves that the plane doesn't exist.

First, of all, draw some pictures and you'll see that a sphere-cylinder intersection is only planar (it's a circle) when the centerline of the cylinder passes through the center of the sphere. That's not the case, here.
If you want "proof" as opposed to pictures and intuition, just use your parametric equations to compute four points $(x_i,y_i,z_i)$ on the curve. Then compute the following determinant
$$
D = \left|\begin{matrix}
x_1  &  y_1  & z_1 &  1 \\
x_2  &  y_2  & z_2 &  1 \\
x_3  &  y_3  & z_3 &  1 \\
x_4  &  y_4  & z_4 &  1 \\
\end{matrix}\right|
$$
Actually, $\tfrac16D$ is the volume of the tetrahedron having the four points as vertices, so $D$ will be zero if and only if the four points are coplanar. If you pick four points at random, you will almost certainly get a non-zero value for $D$, which will show that the curve is non-planar.
