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A. \[5\lambda - 8\mu = 0\]

B. \[7\lambda - 10\mu = 0\]

C. \[10\lambda - 7\mu = 0\]

D. \[8\lambda - 5\mu = 0\]

Answer
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* Midpoint of a line joining two points \[({x_1},{y_1},{z_1}), ({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]

*Directional ratios of a line joining points \[P({x_1},{y_1},{z_1}),Q({x_2},{y_2},{z_2})\] is given by

\[{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}\]

* Directional cosines of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by

\[\dfrac{{{x_2} - {x_1}}}{{PQ}},\dfrac{{{y_2} - {y_1}}}{{PQ}},\dfrac{{{z_2} - {z_1}}}{{PQ}}\] where \[PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \]

First we draw a triangle \[\vartriangle ABC\], where we draw a median from vertex A to the base BC, the midpoint of BC is D.

We find the coordinates of point D using the formula for midpoint of a line.

Midpoint of a line joining two points \[({x_1},{y_1},{z_1}), ({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]

Here the points are \[B( - 1,3,2), C(\lambda ,5,\mu )\]

Mid-point becomes \[\left( {\dfrac{{ - 1 + \lambda }}{2},\dfrac{{3 + 5}}{2},\dfrac{{2 + \mu }}{2}} \right)\]

So the midpoint of BC is \[D\left( {\dfrac{{ - 1 + \lambda }}{2},4,\dfrac{{2 + \mu }}{2}} \right)\].

Now we look at the line joining the points A and D. Since, we have coordinates of the two points, we can write their directional ratios.

Directional ratios of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by \[{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}\]

Here points are \[A(2,3,5),D(\dfrac{{ - 1 + \lambda }}{2},4,\dfrac{{2 + \mu }}{2})\]

So the directional ratios of AD are \[(\dfrac{{ - 1 + \lambda }}{2} - 2),(4 - 3),(\dfrac{{2 + \mu }}{2} - 5)\]

Taking LCM in the each term

\[

(\dfrac{{ - 1 + \lambda - 4}}{2}),(1),(\dfrac{{2 + \mu - 10}}{2}) \\

(\dfrac{{\lambda - 5}}{2}),(1),(\dfrac{{\mu - 8}}{2}) \\

\]

So the directional ratios are \[\dfrac{{\lambda - 5}}{2},1,\dfrac{{\mu - 8}}{2}\] … (1)

Since, the median through A is equally inclined to the coordinate axes then it means the directional cosines of the line are equal.

We know directional cosines of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by

\[\dfrac{{{x_2} - {x_1}}}{{PQ}},\dfrac{{{y_2} - {y_1}}}{{PQ}},\dfrac{{{z_2} - {z_1}}}{{PQ}}\] where \[PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \]

Since, the directional cosines are equal we can write

\[\dfrac{{{x_2} - {x_1}}}{{PQ}} = \dfrac{{{y_2} - {y_1}}}{{PQ}} = \dfrac{{{z_2} - {z_1}}}{{PQ}}\]

Since the length PQ is same in all denominators, we can remove it

\[{x_2} - {x_1} = {y_2} - {y_1} = {z_2} - {z_1}\] … (2)

From equation (2) we can say the directional ratios of lines are equal, So we equate the directional ratios from equation (1).

\[ \Rightarrow \dfrac{{\lambda - 5}}{2} = 1 = \dfrac{{\mu - 8}}{2}\]

Taking first two terms we get

\[ \Rightarrow \dfrac{{\lambda - 5}}{2} = 1\]

Cross multiplying the values

\[

\Rightarrow \lambda - 5 = 1 \times 2 \\

\Rightarrow \lambda - 5 = 2 \\

\]

Shift all constants to one side of the equation

\[

\Rightarrow \lambda = 2 + 5 \\

\Rightarrow \lambda = 7 \\

\]

Taking last two terms we get

\[ \Rightarrow \dfrac{{\mu - 8}}{2} = 1\]

Cross multiplying the values

\[

\Rightarrow \mu - 8 = 1 \times 2 \\

\Rightarrow \mu - 8 = 2 \\

\]

Shift all constants to one side of the equation

\[

\Rightarrow \mu = 2 + 8 \\

\Rightarrow \mu = 10 \\

\]

Now when we take ratio of the values \[\lambda ,\mu \]

\[ \Rightarrow \dfrac{\lambda }{\mu } = \dfrac{7}{{10}}\]

Cross multiplying the values

\[ \Rightarrow 10\lambda = 7\mu \]

Shift all values to one side of the equation

\[ \Rightarrow 10\lambda - 7\mu = 0\]