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Hint: Plane is perpendicular to the line, so the direction ratios of the plane is equal to the direction ratios of the line.

Line passing through the points $\left( {1,2,3} \right)$ and $\left( {3,4,5} \right)$

So the midpoint of the line be$\left( {\dfrac{{1 + 3}}{2},\dfrac{{2 + 4}}{2},\dfrac{{3 + 5}}{2}} \right)$

$ \Rightarrow \left( {\dfrac{4}{2},\dfrac{6}{2},\dfrac{8}{2}} \right) = \left( {2,3,4} \right)$

The direction ratios of the line be $\left( {\left( {3 - 1} \right),\left( {4 - 2} \right),\left( {5 - 3} \right)} \right) = \left( {2,2,2} \right)$

Now, it is given that the plane bisects the line so the plane is passing through midpoint of the line and plane is also perpendicular to the line

Therefore the direction ratios of the plane is equal to the direction ratios of the line.

Let the direction ration of the plane be $\left( {l,m,n} \right)$

$ \Rightarrow \left( {l,m,n} \right) = \left( {2,2,2} \right)$

So the equation of the plane passing through midpoint of the plane $\left( {2,3,4} \right)$ having direction ratios $\left( {l,m,n} \right)$ be

$l\left( {x - 2} \right) + m\left( {y - 3} \right) + n\left( {z - 4} \right) = 0$

Now substitute the value of $\left( {l,m,n} \right)$

$

2\left( {x - 2} \right) + 2\left( {y - 3} \right) + 2\left( {z - 4} \right) = 0 \\

\Rightarrow x + y + z - 9 = 0 \\

\Rightarrow x + y + z = 9 \\

$

So, this is the required equation of the plane.

Note: In such types of questions first find out the midpoint of the line then find out the direction ratios of the line, then remember the key concept that if the plane is perpendicular to the line then the direction ratios of the plane is equal to the direction ratios of the line, then write the equation passing through the midpoint having direction ratios same as line we will get the required answer.

Line passing through the points $\left( {1,2,3} \right)$ and $\left( {3,4,5} \right)$

So the midpoint of the line be$\left( {\dfrac{{1 + 3}}{2},\dfrac{{2 + 4}}{2},\dfrac{{3 + 5}}{2}} \right)$

$ \Rightarrow \left( {\dfrac{4}{2},\dfrac{6}{2},\dfrac{8}{2}} \right) = \left( {2,3,4} \right)$

The direction ratios of the line be $\left( {\left( {3 - 1} \right),\left( {4 - 2} \right),\left( {5 - 3} \right)} \right) = \left( {2,2,2} \right)$

Now, it is given that the plane bisects the line so the plane is passing through midpoint of the line and plane is also perpendicular to the line

Therefore the direction ratios of the plane is equal to the direction ratios of the line.

Let the direction ration of the plane be $\left( {l,m,n} \right)$

$ \Rightarrow \left( {l,m,n} \right) = \left( {2,2,2} \right)$

So the equation of the plane passing through midpoint of the plane $\left( {2,3,4} \right)$ having direction ratios $\left( {l,m,n} \right)$ be

$l\left( {x - 2} \right) + m\left( {y - 3} \right) + n\left( {z - 4} \right) = 0$

Now substitute the value of $\left( {l,m,n} \right)$

$

2\left( {x - 2} \right) + 2\left( {y - 3} \right) + 2\left( {z - 4} \right) = 0 \\

\Rightarrow x + y + z - 9 = 0 \\

\Rightarrow x + y + z = 9 \\

$

So, this is the required equation of the plane.

Note: In such types of questions first find out the midpoint of the line then find out the direction ratios of the line, then remember the key concept that if the plane is perpendicular to the line then the direction ratios of the plane is equal to the direction ratios of the line, then write the equation passing through the midpoint having direction ratios same as line we will get the required answer.

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