Let A(2,3,5), B(-1,3,2), C(lambda, 5, mu) be | vedclass

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(C) Let $M$ be the midpoint of $BC$. The coordinates of $M$ are given by $(\frac{-1+\lambda}{2}, \frac{3+5}{2}, \frac{2+\mu}{2}) = (\frac{\lambda-1}{2

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$10 \lambda - 7 \mu = 0$

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(C) Let $M$ be the midpoint of $BC$. The coordinates of $M$ are given by $(\frac{-1+\lambda}{2}, \frac{3+5}{2}, \frac{2+\mu}{2}) = (\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2})$.The median through $A$ is the line segment $AM$. The direction ratios of $AM$ are $(\frac{\lambda-1}{2} - 2, 4 - 3, \frac{\mu+2}{2} - 5) = (\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2})$.Since the median is equally inclined to the coordinate axes,the direction ratios must be equal,i.e.,$\frac{\lambda-5}{2} = 1 = \frac{\mu-8}{2}$.From $\frac{\lambda-5}{2} = 1$,we get $\lambda - 5 = 2$,so $\lambda = 7$.From $\frac{\mu-8}{2} = 1$,we get $\mu - 8 = 2$,so $\mu = 10$.We need to check the options for the relation between $\lambda$ and $\mu$. Substituting $\lambda = 7$ and $\mu = 10$ into the options:Option $A: 5(7) - 8(10) = 35 - 80 
eq 0$.Option $B: 8(7) - 5(10) = 56 - 50 
eq 0$.Option $C: 10(7) - 7(10) = 70 - 70 = 0$.Thus,$10 \lambda - 7 \mu = 0$ is the correct relation.

Let $A(2,3,5), B(-1,3,2), C(\lambda, 5, \mu)$ be the vertices of $\triangle ABC$. If the median through the vertex $A$ is equally inclined to the coordinate axes,then

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