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Find the equation of the plane parallel to the plane determined by points A, B and C, and passing through the point D

$A(0,0,0)$

$B(1,2,3)$

$C(-3,0,0)$

$D(-1,2,4)$

$\overrightarrow{AB} = <1,2,3>$

$\overrightarrow{AC} = <-3,0,0>$

Since $\overrightarrow{AB} \times \overrightarrow{AC} = -9j + 6k$

Is it true that equation of the plane parallel is:

$-9(y-2)+6(z-4)=0$

$-9y+6z=6$

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The equation of the plan is given by ${\bf n}\cdot ({\bf x}-{\bf x_0})=0$

Since you found that ${\bf n}=<0,-9,6>$ the equation of the plane become $$<0,-9,6>\cdot (<x,y,z>-(-1,2,4))=-9y+6z-6=0$$

That is $-9y+6z=6$ is the correct answer.

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    $\begingroup$ Thanks for confirming my answer! I was quite sure I had done something wrong but feels good to know that I didn't. $\endgroup$ – hax0r_n_code Oct 31 '15 at 17:37
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the normal is $-9ck+6cj$, $c$ not $0$. Then the equation is $-9c(y-2)+6c(z-4)=0$. Now plug in $D$ to determine $c$.

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  • $\begingroup$ I'm sorry but I'm confused by your comments. What is c? $\endgroup$ – hax0r_n_code Oct 31 '15 at 17:34

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