The equation of the plane which bisects the l | vedclass

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(C) Let the points be $A(-1, 2, 3)$ and $B(3, -5, 6)$.The normal vector $\vec{n}$ to the plane is the vector $\vec{AB} = (3 - (-1), -5 - 2, 6 - 3) = (

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$4x - 7y + 3z = 28$

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(C) Let the points be $A(-1, 2, 3)$ and $B(3, -5, 6)$.The normal vector $\vec{n}$ to the plane is the vector $\vec{AB} = (3 - (-1), -5 - 2, 6 - 3) = (4, -7, 3)$.The midpoint $M$ of the line segment $AB$ is $\left( \frac{-1+3}{2}, \frac{2-5}{2}, \frac{3+6}{2} \right) = \left( 1, -\frac{3}{2}, \frac{9}{2} \right)$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting the values: $4(x - 1) - 7(y + \frac{3}{2}) + 3(z - \frac{9}{2}) = 0$.$4x - 4 - 7y - \frac{21}{2} + 3z - \frac{27}{2} = 0$.$4x - 7y + 3z - 4 - \frac{48}{2} = 0$.$4x - 7y + 3z - 4 - 24 = 0$.$4x - 7y + 3z = 28$.

The equation of the plane which bisects the line joining the points $(-1, 2, 3)$ and $(3, -5, 6)$ at a right angle is

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