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Point $ P(5,0,2)$ and line : $x = 1+3t, y = 4 - 2t, -3 + t$ both lie on the same plane. Should I try to find a normal vector to put this in the form $a_1(x - x_1) + a_2(y - y_1) + a_3(z - z_1) = 0 $ ?

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  • $\begingroup$ If $Q(1, 4, -3)$ and $v=(3, -2, 1)$, then $PQ\wedge v$ should be a normal vector. $\endgroup$ Sep 18 '17 at 21:43
  • $\begingroup$ That’s certainly one way to proceed. $\endgroup$
    – amd
    Sep 19 '17 at 0:37
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You have $p = (5,0,2)$ and $r(t) = (1+3t,4-2t,-3+t)$ both lie in a plane $P$. It is clear that $p$ is not on this line and so you can use it to get two other points say $r(t_1)$ and $r(t_2)$. Now you have three points in a plane. Can't you make linearly independent vectors from these, construct a normal and get a plane equation from there?

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