A plane passes through the point A(2, 1, -3). | vedclass

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(A) Let the equation of the plane passing through the point $A(2, 1, -3)$ be $a(x - 2) + b(y - 1) + c(z + 3) = 0$,where $\vec{n} = (a, b, c)$ is the n

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$2x + y - 3z = 14$

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(A) Let the equation of the plane passing through the point $A(2, 1, -3)$ be $a(x - 2) + b(y - 1) + c(z + 3) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane.The distance $d$ of the plane from the origin $(0, 0, 0)$ is given by $d = \frac{|-2a - b + 3c|}{\sqrt{a^2 + b^2 + c^2}}$.To maximize the distance $d$,the normal vector $\vec{n}$ must be parallel to the position vector $\vec{OA} = (2, 1, -3)$.Thus,we set $(a, b, c) = (2, 1, -3)$.The equation of the plane becomes $2(x - 2) + 1(y - 1) - 3(z + 3) = 0$.Expanding this,we get $2x - 4 + y - 1 - 3z - 9 = 0$,which simplifies to $2x + y - 3z = 14$.

$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is

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