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Since the direction vector of the line L is [-4,-1,5] and the normal vector of the plane is [9,-16,4], then by taking the dot product of the direction vector and the normal vector, we get:
`n cdot m = (9)(-4)+(-16)(-1)+(4)(5)`
so the normal vector is perpendicular to the direction vector. This means that the line is either parallel to the plane or in the plane. By substituting the equation of the line into the equation of the plane, we find that
`9(1-4t)-16(-2-t)+4(-3+5t)=26` now solve for t
`9-36t+32+16t-12+20t=26` isolate t on LS
which means that the line is contained in the plane.
The plane contains L.
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