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The full question actually reads "At the point where $t=-1$, find an equation for the plane perpendicular to the line $x=5-3t, y=5t-7, z=-6t$."

I'm not too sure how to use the information $t=-1$. When $t=-1, $ the line is at the point $(8,-13,6)$

I guess this means the plane has to pass through the point $(8,-13,6)$?

If that's the case, then we can read off a vector perpendicular to the line by looking at its equation. We find that $<-3,5,6>$ is perpendicular to the line, so that acts as the slope of our plane. We can turn the information we have now into a plane in normal form: $$-3(x-8)+5(y+13)+6(z-6)=0$$

which equals $$-3x + 5y +6z=53$$

But apparently the correct answer is: $$-3x + 5y +6z=120$$

I'm guessing if I went wrong anywhere, it was with my assumption of the point that the plane passes through. But I'm not sure how else to utilize the information given in the question that $t=-1$...any ideas or help is appreciated.

Edit: apologies, there was a typo in the title. I've fixed it now.

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1 Answer

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The plane has the form

$$-3x + 5y -6z = C$$

For $t=-1$, the point on the line is $(5-3t, 5t-7, -6t)=(8,-12, 6)$. Plug it into above equation, we have

$$C = -3(8)+5(-12)-6(6)=-120$$

So, the equation of the plane is,

$$3x - 5y +6z=120 $$

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