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I tried this problem many times but the answer I am getting is not correct.

Find the equation of the plane passing through $(-2,1,1)$, perpendicular to the line which passing through $(5,-1,2)$ and $(0,5,1)$.

The correct answer to this question is $x+2y-3=0$.

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  • $\begingroup$ what Kind of equation can you use? $\endgroup$ – Dr. Sonnhard Graubner Mar 7 '17 at 18:39
  • $\begingroup$ What is your normal vector? The coefficients of the normal vector are the coefficients of x,y,z in the equation of the plane. Adjust $d$ such that you point lies on you plane. $ax_0 + by_0 + cz_0 = d$ $\endgroup$ – Doug M Mar 7 '17 at 18:44
  • $\begingroup$ HINT: for the normal vector of the plane you can use the vector $$[-5;6;-1]$$ and a Point is given by $$[-2,1,1]$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 7 '17 at 18:44
  • $\begingroup$ Got it..Correct answer is -5x+6y-z=5....Thanks for the help! $\endgroup$ – Chintan Mar 7 '17 at 18:52
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the plane has the form $$(\vec x-[-2,1,1])\cdot [-5,6,-1]=0$$

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