Posted by Anita on Monday, April 9, 2012 at 3:56am.
the normal of the given plane is
(3,0,-5) which is then the direction of your line
A vector equation of that line is
r = (x,y,z) = (2,9,-3) + t(3,0,-5) where t is your parameter
x = 2 + 3t ---> t = (x-2)/3
y = 9 + 0t ---> y = 9
z = -3 - 5t --> t = (z+3)/-5
so
(x-2)/3 = (z+3)/-5 , y = 9
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