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"Find the value of $k$ for which the plane $kx + 4y + 2z - 6 = 0$ is parallel to the line $\frac{x-3}{5} = \frac y1 = \frac{z}{-3}$."

So based on this I know the normal to the plane is $(k, 4, 2)$. I also know the direction vector of the line is $(5, 1, -3)$.

I need to find a vector that is perpendicular to the normal, and then somehow use that with the given direction vector to find $k$, but I'm not quite sure how.

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Hint: Compute the dot product $$[k;4;2]\cdot [5;1;-3]$$ This product must be zero, then you will get your $k$

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