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I am calculating the normal vector to the plane $5x+2y+3z=1$

According to the book:

The normal vector $N$ is often normalized to unit length because in that case the equation $$ d = N ⋅Q + D $$ gives the signed distance from the plane to an arbitrary point $Q$. If $d = 0$, then the point $Q$ lies in the plane. If $d > 0$, we say that the point $Q$ lies on the positive side of the plane since $Q$ would be on the side in which the normal vector points.

How to get thenormal vector $N$? Thank you

Updated: the normailize function $$ q= \sqrt{q_0^2+q_1^2+q_2^2+q_3^2} $$

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