find the distance from point Q to line l (matrix)

Q=(0, 1, 0)

line = < x y z > = < 1 1 1 > + t< -2 0 3 >

Formula:
       __
     ||PQ × u||
D = ------------
       ||u||

where u is the direction vector for the line and
P is a point on the line

Using the direction numbers -2, 0, and 4, you know
that the direction vector for the line is

u = < -2, 0, 4 >

To find a point on the line, let t = 0 and obtain

P = (1, 1, 1)

Thus, since Q = (0, 1, 0)
__
PQ = < 0-1, 1-1, 0-1 > = < -1,0,-1 >
                          __
Now get the cross product PQ × u

__       | i  j  k| 
PQ × u = |-1  0 -1| = i(0+0) - j(-4-2) + k(0+0) = -6j
         |-2  0  4|
  __
||PQ × u|| = ||-6j|| = 6
         ____________    ______    __     _
||u|| = Ö(-2)²+0²+ 4² = Ö4+0+16 = Ö20 = 2Ö5
       __                            _
     ||PQ × u||      6       3     3Ö5
D = ------------ = ----- = ---- = -----
       ||u||        2Ö5     Ö5      5

Edwin