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Find the plane through the point $(1,4,5)$ and perpendicular to the vector $(7,1,4)$.

Answer given to me is $$7x + y + 4z = 31 \text{ (I get this).}$$

and $$r = (1,4,5) + s(4,0,-7) + t(0,4,-1)$$ (I tried but couldn't get the middle vector with 's').

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The two vectors are picked such that they are perpendicular to the normal vector $(7,1,4)$. For them to be perpendicular, their dot product is zero.

We can see easily, the dot product of $(4,0,-7)$ and $(7,1,4)$ is $4\cdot 7+0 \cdot 1+(-7)\cdot 4=0$. You can pick any vector that has one element equal to, and another element opposite to the corresponding ones in $(7,1,4)$, switch their positions, and leave the 3rd one $0$. For example, $(-1,7,0), (1,-7,0),(-4,0,7), (0,4,-1),(0,-4,1)$.

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