Find value of $k$ for a plane for which it is parallel to given line

"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.

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