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Question

(B) Let the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane be $P(3, 2, 1)$.The normal vector $\vec{n}$ to the plane is the vector

Vedclass · Accepted Answer

$3x+2y+z=14$

Vedclass · Answer

(B) Let the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane be $P(3, 2, 1)$.The normal vector $\vec{n}$ to the plane is the vector $\vec{OP} = 3\hat{i} + 2\hat{j} + \hat{k}$.The equation of a plane passing through a point $P(x_1, y_1, z_1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.Substituting the values $a=3, b=2, c=1$ and $(x_1, y_1, z_1) = (3, 2, 1)$:$3(x - 3) + 2(y - 2) + 1(z - 1) = 0$$3x - 9 + 2y - 4 + z - 1 = 0$$3x + 2y + z - 14 = 0$$3x + 2y + z = 14$.Thus,the correct option is $B$.

If the foot of the perpendicular drawn from the origin to the plane is $(3, 2, 1)$,then the equation of the plane is

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