Which point on a diffuse reflecting plane is brightest to the viewer? Given is the plane y = 0. A point light source is positioned at (1,10,1). The 
plane reflects only diffuse. The viewer is positioned at (4,6,4) and sees all 
points on the plane with 0 <= x <= 8 and 0 <= z <= 8. At which point on the 
plane does the viewer perceive the highest reflection?

I think the answer is (1, 0, 1), the projection of the point light source onto the reflecting plane. Because a diffuse reflection scatters light equally in all directions so it doesn't matter where the reflection is relative to the viewer. All that matters is where the reflection is relative to the point light source, because the point light source shines a varying amount of intensity per unit area on different parts of the plane. But the highest concentration of intensity will be shined along the plane's normal vector aka the nearest point on the plane. Is this correct and if not how do you do this? Thank you
 A: Yes, your analisys is right.  Note that the question is not about which point the viewer receives the most light power from, but which one appears brightest.  The "appear brightest" part factors out distance to the viewer.  Distance will only make a object appear smaller, not different in brightness.  Therefore, as you said, the part of the plane that appears brightest is the brightest part of the plane, and will appear to all observers to be the brightest part of the plane.
Added:
From the comments, it is clear people are having a problem with "observed brightness" as apposed to the amount of light reaching the observer.  To be clear, I am talking about the brightness of the object focused on some image plane, like the film or sensor in a camera or your retina.  The light power reaching the observer falls off with the square of the distance.  However, the linear dimension of the focused object on the image plane falls off with distance, which means its area falls off with the square of the distance.  Since the total light hitting the observer also falls off with the square of the distance, the focused projection onto the image plane remains the same brightness.  No laws of conservation were violated since the image of the distant object is smaller, so the total light received is correct.
For a more intuitive feel for this, think of photographing a outdoor scene in daylight.  You have a person 10 feet away and a mountain 10 miles away, both illuminated the same by the sun.  Even though the mountain is 5000 time further away, the two objects produce about the same brightness on the film.  If this weren't true, all those tourist pictures of someone ovelooking the Grand Canyon would be impossible.
I think the question is rather clearly worded to ask for this apparent visible brightness, not the total light power since that's not how we perceive brightness.
A: Edit 3: Yes, you are correct. For an excellent explanation, see the Wikipedia entry on Lambert's cosine law. You can safely ignore the remainder of this answer. NB: Thanks for the question!

Edit 1: It appears that in this answer I reasoned that we are looking for luminance (even though I was not familiar with the terminology at the time), and having now looked at Wikipedia, I still think that is correct. I assume that what I "derived" below comes fairly close or exactly down to the use of its definition. Olin's answer seems to refer to luminosity, and since I didn't "derive" that, I think that luminosity is "wrong" for this situation. :)
Edit 2: Also, I took your "a diffuse reflection scatters light equally in all directions" for granted and referring to points, although it seemed kind of weird. But since browsing the Wikipedia, I came across Lambert's cosine law, which indicates that I might have to read that differently. Correcting for a Lambertian surface, I fear that Olin might very well have been right all along. (And the question then does become almost tautological, because the cosines cancel out by definition, which is why his answer is so short.)


I'm going to have a half-hearted shot at this. It's a bit messy, and I recommend not relying on it. It also shows how you can make a mess of something that perhaps is as easy as Olin makes it seem. Perhaps this is even plain wrong...
The length of the first leg from $(1,10,1)$ to $(x,0,z)$ equals $$d_1=\sqrt{(1-x)^2+10^2+(1-z)^2}.$$
The intensity of the incoming light at $(x,0,y)$ is proportional to $$\frac{1}{d_1^2}=\frac{1}{(1-x)^2+10^2+(1-z)^2}.$$ Note or simply assume* that nearby points $(x\pm\delta,0,z\pm\varepsilon)$ have virtually equal intensity of incoming light.
The length of the second leg from $(x,0,z)$ to $(4,6,4)$ equals $$d_2=\sqrt{(4-x)^2+6^2+(4-z)^2}.$$
The intensity of light received from this exact point is proportional to $$\frac{1}{d_1^2}\frac{1}{(4-x)^2+6^2+(4-z)^2}=\frac{1}{(1-x)^2+10^2+(1-z)^2}\frac{1}{(4-x)^2+6^2+(4-z)^2}.$$ (I think I am assuming something here, but I'm not sure what.)
But note that as $d_2$ increases, point ($x,0,z$) becomes "smaller". You might interpret this as having to "include more points" $(x\pm\delta,0,y\pm\varepsilon)$ in the immediate vicinity (neighbourhood). But: How much more points?
I suggest that that depends on the angles $\varphi\pm \gamma$ between the line from $(4,6,4)$ to $(x\pm\delta,0,z\pm\varepsilon)$ and the $xz$-plane (with $\delta$ and $\varepsilon$ possibly dependent, because they represent a "viewing cone").
So the amount of points to be included is proportional to the area consisting of all intersections of all lines from $(4,6,4)$ to some $(x\pm\delta,0,z\pm\varepsilon)$ and the $xz$-plane with angles $\varphi\pm\gamma$. Let's call this $A(x,0,z;\gamma)$, because I'm very bad at trigonometry.
So, the final intensity to be perceived is proportional to $$A(x,0,z;\gamma)\frac{1}{(1-x)^2+10^2+(1-z)^2}\frac{1}{(4-x)^2+6^2+(4-z)^2}.$$
Maximize that with respect to $(x,0,z)$. This should give you an answer in terms of $\gamma$.
Take a realistic $\gamma$ or take the limit $\gamma\downarrow0$. (Taking the limit will probably justify the earlier assumption*.)
