Let's say your three points are $\vec{p}_1 = ( x_1 , y_1 , z_1 )$, $\vec{p}_2 = ( x_2 , y_2 , z_2 )$, and $\vec{p}_3 = ( x_3 , y_3 , z_3 )$.
Let
$$\vec{n} = \left ( \vec{p}_2 - \vec{p}_1 \right ) \times \left ( \vec{p}_3 - \vec{p}_1 \right )$$
i.e.
$$\begin{cases}
x_n = (y_2 - y_1) (z_3 - z_1) - (z_2 - z_1) (y_3 - y_1) \\
y_n = (z_2 - z_1) (x_3 - x_1) - (x_2 - x_1) (z_3 - z_1) \\
z_n = (x_2 - x_1) (y_3 - y_1) - (y_2 - y_1) (x_3 - x_1) \end{cases} $$
If the three points are collinear, then $\vec{n} = ( 0 , 0 , 0 ) = 0$. Otherwise, the three points are on a plane, with $\vec{n}$ being normal (perpendicular) to the plane.
As mentioned in a comment, if we look at the triangle from the side the normal vector $\vec{n}$ points to, the points are in counterclockwise order; but, if we look at the triangle from the other side, they are in clockwise order.
If we know that the three points are on a plane with normal vector $\vec{k}$, then
$$\begin{cases}
\vec{k} \cdot \vec{n} = \vec{k} \cdot \left ( \left ( \vec{p}_2 - \vec{p}_1 \right ) \times \left ( \vec{p}_3 - \vec{p}_1 \right ) \right ) \gt 0, & \text{ counterclockwise } \\
\vec{k} \cdot \vec{n} = \vec{k} \cdot \left ( \left ( \vec{p}_2 - \vec{p}_1 \right ) \times \left ( \vec{p}_3 - \vec{p}_1 \right ) \right ) \lt 0, & \text{ clockwise } \\
\vec{k} \cdot \vec{n} = \vec{k} \cdot \left ( \left ( \vec{p}_2 - \vec{p}_1 \right ) \times \left ( \vec{p}_3 - \vec{p}_1 \right ) \right ) = 0, & \text{ oops } \end{cases}$$
The "oops" case covers several possible situations. For example, if the three points are collinear, then $\vec{n} = 0$. Or, if the three points are on a plane parallel to $\vec{k}$, the result is zero also.
Here is an approach that avoids dividing into cases as well as any trigonometry. Suppose that the directions $d^1 = (d^1_1,d^1_2)$ and $d^2 = (d^2_1,d^2_2)$ are given in clockwise order. Rotate these vectors by $90^\circ$ clockwise and counterclockwise respectively to product
$$
l = (d_2^1,-d^1_1), \quad r = (-d^2_2, d^2_1).
$$
A point $x = (x_1,x_2)$ will lie within the cone if and only if it satisfies $x \cdot l \geq 0$ and $x \cdot r \geq 0$, where $v \cdot w$ denotes the dot-product of vectors $v$ and $w$. More specifically, we have $x \cdot l \geq 0$ iff $x$ lies to the "right" of the "left-side" boundary, and $x \cdot r \geq 0$ iff $x$ lies to the "left" of the "right-side" boundary.
We are given two endpoints $p^1 = (p^1_1,p^1_2)$ and $p^2 = (p^2_1,p^2_2)$. The line connecting these points is the set of all points
$$
p(t) = (1-t)p^1 + tp^2
$$
with $t \in \Bbb R$. Note that $p(t)$ is on the line segment connecting the two points when $0 \leq t \leq 1$. Moreover, $p(0) = p^1$ and $p(1) = p^2$.
We now find the "times" $t$ at which this line crosses either of the boundaries. That is, we solve
$$
l \cdot p(t_l) = 0 \implies (1-t_l)(l \cdot p^1) + t_l(l \cdot p^2) = 0 \implies t_l = \frac{l \cdot p^1}{(l \cdot p^1) - (l \cdot p^2)},\\
l \cdot p(t_r) = 0 \implies (1-t_r)(l \cdot p^1) + t_r(l \cdot p^2) = 0 \implies t_r = \frac{r \cdot p^1}{(r \cdot p^1) - (r \cdot p^2)}.
$$
If either of these numbers satisfy $0 \leq t \leq 1$, plug into $p(t)$ to produce the associated point.
The only case not accounted for here is division by zero, which occurs when the line segment is parallel to one of the boundaries.
Best Answer
Take any point $P$ in your 3D-space, and if we call the given points $A_1,\ldots,A_n$, calculate the sum of the vector products $\vec{S} = \frac{1}{2}\sum_{i=1}^n{\overrightarrow{PA_i}\times\overrightarrow{PA_{i+1}}}$, where indices are taken modulo $n$. You can easily verify that the value of $\vec{S}$ does not depend on the choice of $P$.
So we can assume that $P$ lies in your plane, so each cross product from the sum has both factors as vectors in your plane, so each cross product is normal to your plane. Hence the sum is also a vector that is normal to your plane. The length of $\vec{S}$ calculates the area of your polygon. The direction of $\vec{S}$ shows you the direction of the normal around which the points $A_1,\ldots,A_n$ would be considered counter-clockwise.
This means you need to calculate $\vec{S}$ (or $2\vec{S}$, as the value of the area is not important here), with any choice of $P$ (usually you just use the origin of your coordinate system), then you need to compare $\vec{S}$ with your given plane normal $\vec{n}$: Are they pointing in the same or opposite directions? This check can be simply done with a simple dot product: Is $\vec{S} \cdot \vec{n}$ positive (same direction) or negative (opposite direction)?