Question 907680
Use the distance formula.
{{{D^2=(-2-3t)^2+(2-2t)^2}}}
{{{D^2=(9t^2+12t+4)+(4t^2-8t+4)}}}
{{{D^2=13t^2+4t+8}}}
To minimize the distance squared, take the derivative of the distance squared and set it equal to zero.
{{{d(D^2)/dt=26t+4}}
{{{26t+4=0}}}
{{{26t=-4}}}
{{{t=-4/26}}}
{{{t=-2/13}}}
{{{drawing(300,300,-3,3,-3,3,grid(1),circle(-2,2,0.2),circle(-6/13,-4/13,0.2),blue(line(-2,2,-6/13,-4/13)))}}}
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You could also do it geometrically.
(3t,2t) defines a line through the origin where {{{y=(2/3)x}}}
You could find the line perpendicular to that line through the point (-2,2).
Perpendicular lines have negative reciprocal slopes so the slope would be {{{m=-3/2}}}
Using the point slope form of a line,
{{{y-2=-(3/2)(x-(-2))}}}
{{{y-2=-(3/2)x-3}}}
{{{y=-(3/2)x-1}}}
So then finding the intersection of the two lines, 
{{{(2/3)x=-(3/2)x-1}}}
{{{x(2/3+3/2)=-1}}}
{{{x(13/6)=-1}}}
{{{x=-6/13}}}
So then,
{{{3t=-6/13}}}
{{{t=-2/13}}}
{{{drawing(300,300,-3,3,-3,3,grid(1),circle(-2,2,0.2),circle(-6/13,-4/13,0.2),circle(-6/13,-4/13,0.2),graph(300,300,-3,3,-3,3,(2/3)x,-(3/2)x-1))}}}