# Find the intersection of the line and plane: x+2y+2z=3, r(t)=<2,-3,3>+t<1,3,-3>

Find the intersection of the line and plane:
$x+2y+2z=3\phantom{\rule{0ex}{0ex}}r\left(t\right)=<2,-3,3>+t<1,3,-3>$
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Raina Russo
Equation of plane is:
$x+2y+2z=3$
and Equation of Straight line is:
$r\left(t\right)=<2,-3,3>+t<1,3,-3>$
The cantesian equation of straight line

Let the point of intersection be
$P\left(2+t,3t-3,3-3t\right)$
So:
$P\left(2+t,3t-3,3-3t\right)\phantom{\rule{0ex}{0ex}}$
Since the point P lies on the plane
From $x+2y+27=3\phantom{\rule{0ex}{0ex}}⇒\left(2+t\right)+2\left(3t-3\right)+2\left(3-3t\right)=3\phantom{\rule{0ex}{0ex}}⇒2+t+6t-6+6-6t=3\phantom{\rule{0ex}{0ex}}t=3-2\phantom{\rule{0ex}{0ex}}t=1$
put $t=1\phantom{\rule{0ex}{0ex}}x=2+1\phantom{\rule{0ex}{0ex}}=3$
The point of intersection
$P=\left(3,0,0\right)$