The equation of the plane, which bisects the | vedclass

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(C) Let the points be $A(1, 2, 3)$ and $B(3, 4, 5)$.The midpoint $M$ of the line segment $AB$ is given by $M = \left(\frac{1+3}{2}, \frac{2+4}{2}, \fr

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$x+y+z=9$

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(C) Let the points be $A(1, 2, 3)$ and $B(3, 4, 5)$.The midpoint $M$ of the line segment $AB$ is given by $M = \left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) = (2, 3, 4)$.The direction ratios of the line segment $AB$ are $(3-1, 4-2, 5-3) = (2, 2, 2)$.Since the plane bisects $AB$ at right angles, the line $AB$ is normal to the plane. Thus, the normal vector is $\vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k}$, which can be simplified to $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.The equation of a plane passing through a point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Here, $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.Substituting these values, we get: $((x\hat{i} + y\hat{j} + z\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.$(x-2)\hat{i} + (y-3)\hat{j} + (z-4)\hat{k} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.$(x-2) + (y-3) + (z-4) = 0$.$x + y + z - 9 = 0$, or $x + y + z = 9$.

The equation of the plane, which bisects the line joining the points $(1, 2, 3)$ and $(3, 4, 5)$ at right angles is

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