Planarity of a 3D curve defined parametrically If you have a 3D curve defined parametrically by $t$ for example:
$$P=\cos\left(t+\frac{\pi}{4}\right) \frac{3}{2} {\bf i} + 3 \sin\left(t+\frac{\pi}{4}\right) {\bf j} + \cos\left(t+\frac{\pi}{4}\right) \frac{3\sqrt3}{2} {\bf k}$$
How do you determine if it lies in a certain plane?
 A: If you have the plane normal direction $\bf n$ and a point on the plane $\bf r$ then the point $\bf p$ lies on the plane if
$$ {\bf n} \cdot ( {\bf p}-{\bf r} ) =0$$
A: A general answer is that a necessary and sufficient condition for a curve to be planar is that its torsion, given by the following formula (https://en.wikipedia.org/wiki/Torsion_of_a_curve)
$$\tau=\dfrac{\det(P'(t),P''(t),P''(t))}{\|P'(t) \times P'''(t)\|}$$
is zero. (under the condition that the third derivative exists...), itself reducible to the simpler criterion:


$$\tag{1} \det(P'(t),P''(t),P'''(t))=0.$$


For the example you give, we easily check that condition (1) is fulfilled. In fact, setting $T=t-\pi/4$ and factorizing the coefficients, we obtain:
$$\left|\begin{array}{rrr}-\sin(T)&-\cos(T)&\sin(T)\\ 
\cos(T)&-\sin(T)&-\cos(T)\\
-\sin(T)&-\cos(T)&\sin(T)\end{array}\right|$$
which is $0$ because rows 1 and 3 are identical.
But in this case, there is a simpler way. 
It suffices to obseerve the following implication 
$$\begin{cases}
x= \frac{3}{2}\cos(t+\frac{\pi}{4})\\
y=3 \sin(t+\frac{\pi}{4})\\
z= \frac{3\sqrt3}{2}\cos(t+\frac{\pi}{4})
\end{cases} \ \ \ \implies \ \ \ z=\sqrt{3} \  y$$
which is translated into set langage by:
"The curve is included into the plane with equation  $z=\sqrt{3} \  x.$"
Moreover, this curve is described by a periodical parametrization (see figure below).

