‖PQ×u‖
D = --------
      ‖u‖

where P is the point P(2,-5,-2) and Q is any point on the line, say when t=0

 x=4+2t, y=-4-3t, z=5+4t  we have the point Q(4,-4,5), 

and u =  < 2, -3, 4>, the direction vector for the line.

We calculate the vector PQ = < 4-2, -4-(-5), 5-(-2) > = < 2, 1, 7 >

We find the cross product:

PQ×u =  = 25i + 6j - 8k

We find the norms:

‖PQ×u‖ = Ö25²+6²+(-8)² = Ö725  = 5Ö29

‖u‖ = Ö2²+(-3)²+4² = Ö29 
 
     ‖PQ×u‖     5Ö29
D = -------- = ------ = 5 
      ‖u‖        Ö29



Edwin