To find the normal vector of a plane, use the point normal equation.

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Direction vector of straight line S = (5,2, 1) passing through point Q (4 4,3,0). Vector PQ=( 1,-4,2)

The normal vector of the plane is perpendicular to both S and PQ, so the normal vector can be taken as s×PQ=(6, -9, -22).

So the equation of the plane is 8(x-3)-9(y- 1)-22(z+2)=0, that is, 8x-9y-22z-59=0.