Intersect the circle with the two lines and the circle forming the sector's boundary; in each case, if there are intersections, check whether they lie within the part (line segment or arc) forming the boundary. If you don't find intersections, the sector and circle are either disjoint, or one is contained in the other; you can check for the latter case by determining whether the centre of either circle is inside the other object.
[Edit in response to the comment:]
The sector is bounded by two line segments and a circular arc, which are parts of two lines and a circle, respectively. To intersect a line given by $\vec n\cdot\vec x=c$ (with $\vec n$ a unit normal vector) and a circle given by $(\vec x-\vec x_0)^2=r^2$, find the (signed) distance $d=\vec n\cdot\vec x_0-c$ of the circle's origin from the line. If $d\gt r$ there are nor intersections; if $d\le r$, the intersections are at
$$\vec x_0-d\vec n\pm\sqrt{r^2-d^2}\,\vec y\;,$$
where $y$ is a unit vector with $\vec n\cdot\vec y=0$, i.e. in either direction along the line.
To intersect two circles given by $(\vec x-\vec x_1)^2=r_1^2$ and $(\vec x-\vec x_2)^2=r_2^2$, subtract the two equations to obtain
$$2(\vec x_2-\vec x_1)\cdot\vec x=r_1^2-r_2^2-\vec x_1^2+\vec x_2^2\;,$$
which is an equation for a line perpendicular to the line connecting the two centres, which you can intersect with one of the circles as described above to find the intersections of the circles.
By equidistant if you mean that the shortest distance between any pair of these points is the same then there are an infinite number of correct answers. For example $(0,0,500),(0,-250\sqrt{3},-250),(0,250\sqrt{3},-250)$ lie on a great circle and are equidistant.
Best Answer
Taking your example of $R=100$ and $d=1$, you get a central angle of
$$\alpha=2\arcsin\frac{d}{2R}\approx0.573°\approx\frac{2\pi}{628}$$
So your points wouls most closely resemble a regular $628$-gon. But not exactly: the $628$-gon has $d\approx1.0005$ while the $629$-gon has $d\approx0.9989$. Therefore you can't place points at unit distance from one another along a circle of radius $100$ all the way around. Not exactly.
But perhaps you only want points on a part of the circle, so they don't have to get back exactly to their starting point. Or perhaps a slight deviation in radius or distance is acceptable. Then you can get coordinates for your points using
$$ x_k = R\cos(k\alpha) \qquad y_k = R\sin(k\alpha) $$
for $k\in\{0,1,2,\dots,n-1\}$. If you aim for an exact $n$-gon, you'd use $\alpha=\frac{2\pi}n$ while for exact lengths $R$ and $d$ but a non-closing sequence you'd use the angle as computed above.