Cube question help.

Here I wrote in the info given in the problem and I drew point S according to the directions in the hint.

Let's just look at triangle DRS by itself:

If only we knew what SA was, we could find RD!

Well, let's look at triangle DCS by itself:

Ah, now we can find SA!

$\frac{5}{5+SA}=\frac{2}{SA} \\~\\ (5)(SA)=10+(2)(SA) \\~\\ SA=\frac{10}{3}$

Now looking back at triangle DRS:

$\frac{DR}{5+\frac{10}{3}}=\frac{1}{\frac{10}{3}} \\~\\ DR*\frac{3}{25}=\frac{3}{10} \\~\\ DR=\frac{3}{10}*\frac{25}{3}=\frac{5}{2}$

Let  B  = (0,0, 0)      Let C  = (0, 5, 0)     Let  P = ( 3, 0 ,0)    Let D  = (5,5, 0)    Let Q  = ( 5,,0 , 1)

Equation of CP     y = (-5/3)x + 5  →   [ y - 5] / (-5/3)   = x

Equation of AD  =   x  = 5

Y Intersection  of   CP, AD      [ y - 5 ] / (-5/3)   = 5  →   y - 5  = 5 (-5/3)   →  y =  -25/3 + 5   =   -10/3

So "S"   =   (5, -10/3, 0 )

So AS  = 10/3 

Now, the plane apex intersects AD at S.....thus, SRD forms a triangle  such that  triangle SRD  ≈ SQA

And  QA/AS   ≈ RD / DS

1 / ( 10/3)   =  RD/ (5 + 10/3)

3/10  = RD / ( 25/3)

(3/10) (25/3)   = RD   = 25 / 10   = 2.5

Check  ...distance from S to R  ≈  8.7  = the hypotenuse of SRA

And  DR^2 + SD^2  =   sqrt [ (5 + 10/3)^2  + 2.5^2 ] ≈  8.7=  SR

Hey.....hectictar  and I got the same answer....albeit, with slightly different approaches........his is a little better, IMHO, but two great minds couldn't be wrong.....LOL!!!!!

Let ABCDEFGH be a cube of side length 5, as shown.

Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1.

The plane through C, P, and Q intersects line DH at R. Find DR. 

$Let\ \vec{C} = \langle 0,0,0 \rangle\\ Let\ \vec{P} = \langle 5,3,0 \rangle\\ Let\ \vec{Q} = \langle 5,5,1 \rangle\\ Let\ DR = z \\ Let\ \vec{R} = \langle 0,5,z \rangle$

The plane through C, P, and Q:

$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{C} + s \cdot(\vec{P}-\vec{C}) + t \cdot(\vec{Q}-\vec{C}) \quad & | \quad \vec{x} = \vec{R} \quad \vec{C} = \vec{0} \\ \vec{R} &=& \vec{0} + s \cdot(\vec{P}-\vec{0}) + t \cdot(\vec{Q}-\vec{0}) \\ \vec{R} &=& s \cdot \vec{P} + t \cdot \vec{Q} \quad & | \quad \vec{P} =\begin{pmatrix}5 \\ 3 \\ 0\end{pmatrix} \quad \vec{Q}=\begin{pmatrix}5 \\ 5 \\ 1\end{pmatrix} \quad \vec{R}=\begin{pmatrix}0 \\ 5 \\ z\end{pmatrix} \\ \begin{pmatrix}0 \\ 5 \\ z\end{pmatrix} &=& s \cdot \begin{pmatrix}5 \\ 3 \\ 0\end{pmatrix} + t \cdot \begin{pmatrix}5 \\ 5 \\ 1\end{pmatrix} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (1) & 0&=& 5s+5t \\ & 0&=& s+ t \\ & s&=& -t \\\\ (2) & 5&=& 3s+5t \\ & 5&=& 3(-t) + 5t \\ & 5&=& 2t \\ & t&=& \frac{5}{2} \\\\ (3) & z&=& 0\cdot s + t \\ & z&=& t \\ & z&=& \frac{5}{2} \\ \hline \end{array}$

DR = $\frac{5}{2}$