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

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 = 3 and AQ = 1. The plane through C, P, and Q intersects line DH at R. Find DR.

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∠CPQ ≈ 68.199º

∠DCR ≅ ∠APQ = 18.43494881º

DR = tan(∠DCR) * CD = 1.66666667    or     5/3

Thanks, jugoslav.....here's another way without using trig.....

Let B  = (0,0)   A  = (5,0)   C  = (0,5)   and   D = (5,5)   P  = (2,0)

Slope  of  CP   =  -5/2

Equation of line through  CP .....  y = (-5/2)x + 5

Let S  be the intersection of  the line through segment  RQ  and the line through  CP

The x coordinate of this point   = 5....so we  have

y = -(5/2)(5)  + 5

y=  -25/2  +  5

y  = -15/2

So S = (5, -15/2)

So SQ  =  15/2

Using similar triangles

AQ / SA  =  DR/SD

AQ/ SA = DR  / (SA + AD)

1/ (15/2)  = DR  / ( 15/2  + 5)

(2/15)  = DR / (25/2)

DR = (25/2) (2/15)  = 50/30   = 5/3