graphics – Find point along x axis where yz plane flips to back facing after perspective projection. Have a solution but don’t know why it works

I am trying to find the point along the x axis where a plane with normal pointing along the x axis would flip to back facing after perspective projection. Essentially the red line in this image.

My first attempt to find it I just sampled along the x axis checking if a triangle there was back facing after projection. This only gave me an approximation but I could make the approximation as accurate as I wanted by just sampling more planes so I used this as a baseline to verify other solutions.

I then tried the following:

vec4 center = projection.inverse()*vec4(0,0,0,1);
center /= center.w;
float d = dot(, vec3(1,0,0));
vec3 point = vec3(1,0,0)*d;

But this did not work, it was close but not right. See this image, the red line is the sampled point and the blue is the one calculated from that.

I then tried changing the calculation for center to projection.inverse()*vec4(0,0,-1,1) to see what that would look like. With the the sampled point in red, the blue line being the previous version and the green the new version. I got the following image which as you can see is still not correct.

But I noticed that the green was always directly between the blue and the red so I tried the following.

vec4 c1 = projection.inverse()*vec4(0,0,0,1);
c1 /= c1.w;
float d1 = dot(, vec3(1,0,0));
vec4 c2 = projection.inverse()*vec4(0,0,-1,1);
c2 /= c2.w;
float d2 = dot(, vec3(1,0,0));
float d = (d2-d1)*2 + d1;
vec3 point = vec3(1,0,0)*d;

Which seems to be correct but I have no idea why that works. It is always extremely close to the sampled point so I assume it is correct.

My question is why does this work?