If the position vectors of the points A and B | vedclass

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(C) The normal vector $\vec{n}$ to the plane is along $\overrightarrow{AB}$.$\overrightarrow{AB} = \vec{B} - \vec{A} = (\hat{i}-2 \hat{j}-4 \hat{k}) -

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$2 x+3 y+6 z+28=0$

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(C) The normal vector $\vec{n}$ to the plane is along $\overrightarrow{AB}$.$\overrightarrow{AB} = \vec{B} - \vec{A} = (\hat{i}-2 \hat{j}-4 \hat{k}) - (3 \hat{i}+\hat{j}+2 \hat{k}) = -2 \hat{i}-3 \hat{j}-6 \hat{k}$.Since the plane is perpendicular to $\overrightarrow{AB}$,the normal vector to the plane can be taken as $\vec{n} = 2 \hat{i}+3 \hat{j}+6 \hat{k}$.The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = a \hat{i}+b \hat{j}+c \hat{k}$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the point $B(1, -2, -4)$ and the normal vector components $(2, 3, 6)$:$2(x-1) + 3(y+2) + 6(z+4) = 0$.$2x - 2 + 3y + 6 + 6z + 24 = 0$.$2x + 3y + 6z + 28 = 0$.

If the position vectors of the points $A$ and $B$ are $3 \hat{i}+\hat{j}+2 \hat{k}$ and $\hat{i}-2 \hat{j}-4 \hat{k}$ respectively,then the equation of the plane passing through $B$ and perpendicular to $AB$ is

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