If O is the origin and the coordinates of P a | vedclass

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(C) The origin $O$ is $(0, 0, 0)$ and point $P$ is $(1, 2, -3)$.Since the plane is perpendicular to $OP$,the vector $\vec{OP}$ acts as the normal vect

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$x + 2y - 3z - 14 = 0$

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(C) The origin $O$ is $(0, 0, 0)$ and point $P$ is $(1, 2, -3)$.Since the plane is perpendicular to $OP$,the vector $\vec{OP}$ acts as the normal vector $\vec{n}$ to the plane.$\vec{n} = \vec{OP} = (1 - 0)\hat{i} + (2 - 0)\hat{j} + (-3 - 0)\hat{k} = 1\hat{i} + 2\hat{j} - 3\hat{k}$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ 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 $a=1, b=2, c=-3$ and $(x_1, y_1, z_1) = (1, 2, -3)$:$1(x - 1) + 2(y - 2) - 3(z + 3) = 0$$x - 1 + 2y - 4 - 3z - 9 = 0$$x + 2y - 3z - 14 = 0$.

If $O$ is the origin and the coordinates of $P$ are $(1, 2, -3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$ is

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