I agree with the results you show. For part (a), you would find that the point (we'll call it $ \ E \ $ ) in plane $ \ S \ $ closest to $ \ Q \ $ lies at $ \ ( \frac{26}{6} , \frac{1}{6} ,\frac{19}{6} ) \ , $ the distance separating points $ \ Q \ $ and $ \ E \ $ being $ \ \frac{\sqrt{66}}{6} $ .
There are a few ways to go about finding the rate at which the "shadow" of the particle moving along line $ \ L \ $ traverses plane $ \ S \ $ . We will start with a "naive" method and see that we can develop a more sophisticated approach from that.
The straightforward way would be to determine the point $ \ D \ $ on the plane which is closest to external point $ \ P \ $ . This point can be thought of as the "shadow" of $ \ P \ $ , since the closest point on a plane to an external point lies on along a normal line to the plane, which here is also the direction of the "illumination". As was done to locate point $ \ E \ $ , we set up parametric equations for the normal line to the plane which passes through $ \ P \ $ ,
$$x \ = \ 3 \ + \ 4v \ \ , \ \ y \ = \ 2 \ - \ v \ \ , \ \ z \ = \ -2 \ - \ 7v \ , $$
and solve $ \ 4x - y - 7z + 5 = 0 \ $ for the value of $ \ v \ $ of the point $ \ D \ $ in the plane. Passing over the algebra of the calculation, we obtain $ \ 66v + 29 = 0 \ \Rightarrow \ v = -\frac{29}{66} \ , $ giving us $ \ D \ ( \frac{82}{66} , \frac{161}{66} ,\frac{71}{66} ) \ . $
Applying the "distance formula", the separation in plane $ \ S \ $ between $ \ D \ $ and $ \ E \ $ is
$$ \sqrt{\frac{204^2 \ + \ 150^2 \ + \ 138^2}{66^2}} \ = \ \frac{\sqrt{83160}}{66} \ = \ \sqrt{\frac{1260}{66}} \ \approx \ 4.369 \ . $$
Since this is the distance the particle's shadow moves on the plane in 2 time-units, the speed of the shadow is $ \ \frac{\sqrt{83160}}{132} \ = \ \sqrt{\frac{315}{66}} \ \approx \ 2.185 \ $ .
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We worked matters out that way first for the purpose of checking our other methods. The next thing we can investigate is the motion of the particle in the plane containing $ \ \vec{v} \ $ and the two normal lines $ \ DP \ $ and $ \ EQ \ $ . In a single time-unit, the particle itself travels a distance given by the length of $ \ \vec{v} \ $ , which is $ \ \sqrt{1^2 + (-1)^2 + (-2)^2} \ = \sqrt{6} \ $ . The distance between points $ \ D \ $ and $ \ P \ $ is
$$ \sqrt{\frac{116^2 \ + \ 29^2 \ + \ 203^2}{66^2}} \ = \ \frac{\sqrt{55506}}{66} \ = \ \frac{29}{\sqrt{66}} \ \approx \ 4.369 \ , $$
while the separation between $ \ E \ $ and $ \ Q \ $ was already found to be $ \ \frac{\sqrt{66}}{6} \ = \ \frac{11}{\sqrt{66}} \ . $
Hence, the perpendicular distance of the particle from plane $ \ S \ $ has diminished by $ \ \frac{18}{\sqrt{66}} \ $ in 2 time-units, or $ \ \frac{9}{\sqrt{66}} \ $ per time-unit. Now, this change in distance along the normal direction to the plane forms a right triangle with the change in distance parallel to the plane and the vector $ \ \vec{v} \ , $ which serves as the hypotenuse. So the length of the "leg" parallel to the plane gives the velocity of the particle along the plane, and thus the speed of the particle's shadow is found from
$$ || \ \vec{v} \ ||^2 \ - \ (v_{\perp})^2 \ = \ (v_{\parallel})^2 $$
$$ \Rightarrow \ (v_{\parallel})^2 \ = \ ( \ \sqrt{6} \ )^2 \ - \ ( \ \frac{9}{\sqrt{66}} \ )^2 \ = \ \frac{315}{66} \ \Rightarrow \ v_{\parallel} \ = \ \sqrt{\frac{315}{66}} \ , $$
as before.
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This suggests a still more direct way to the "parallel velocity". Still working in the plane of vector $ \ \vec{v} \ $ and the two normal line segments, we can simply consider the "scalar projection" of $ \ \vec{v} \ $ onto the normal vector to the plane, $ \ \vec{n} \ $ , given by
$$ \frac{\vec{v} \cdot \vec{n}}{|| \ \vec{n} \ ||} \ = \ \frac{\langle 1, -1, 2 \rangle \cdot \langle 4, -1, -7 \rangle}{ \sqrt{4^2+1^2+7^2} } \ = \ \frac{-9}{\sqrt{66}} , $$
which, as we have seen, represents the speed at which the particle is approaching the plane along the normal direction. We complete the calculation as we did previously, using $ \ (v_{\parallel})^2 \ = \ ( \ \sqrt{6} \ )^2 \ - \ ( \ -\frac{9}{\sqrt{66}} \ )^2 \ . $
Note that, in this last method, we have pretty much wrested the problem free of any discussion at all about the points in the plane $ \ S \ $ .
