There are probably neater ways of doing this, but you could find the value

of t which minimises the distance from a point on L to P, then you have a pair

of points on on the desired line from which the required parametric equation

can be derived.

If the distance between points is minimised, so is the square distance.

The square distance from P to L(t) is:

D(t)=(7-2t)^2+(-1-2t)^2+(-3-t)^2

This is minimised by a value of t such that (d/dt)D(t)=0,

Now:

(d/dt)D(t)=2(7-2t)(-2) + 2(-1-2t)(-2) +2(-3-t)(-1)=18t-18,

so if (d/dt)D(t)=0 t=1.

So the two points on the required line are the given point P=(3,1,-2), and

the point L(1)=(-2,4,3).

Then the line M(lambda)=lambda P + (1-lambda) L(1) is the required line,

or:

x=3 lambda - 2 (1-lambda)=-2 + 5 lambda,

y=1 lambda + 4 (1-lambda)= 4 - 3 lambda

z=-2 lambda +3 (1-lambda)= 3 - 5 lambda.

RonL

(you will need to check the algebra here, there is probably a mistake

somewhere but it should give you a idea of how to solve it).