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(A) The equation of a plane passing through a point $P(x_1, y_1, z_1)$ and having a normal vector $\vec{n} = (a, b, c)$ is given by $a(x-x_1) + b(y-y_

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$2x-y+4z-21=0$

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(A) The equation of a plane passing through a point $P(x_1, y_1, z_1)$ and having a normal vector $\vec{n} = (a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$. Here,the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $P(2, -1, 4)$. This means the vector $\vec{OP} = (2, -1, 4)$ is the normal vector to the plane. So,$a = 2, b = -1, c = 4$. The equation of the plane is $2(x-2) - 1(y-(-1)) + 4(z-4) = 0$. $2(x-2) - 1(y+1) + 4(z-4) = 0$. $2x - 4 - y - 1 + 4z - 16 = 0$. $2x - y + 4z - 21 = 0$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

If the foot of the perpendicular drawn from the origin to a plane is $P(2,-1,4)$,then the equation of the plane is

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