ABC is a triangle in a plane with vertices `A(2,3,5),B(-1,3,2)` and `C(lamda,5,mu)`. If the median through A is equally inclined to the coordinate axes, then the value of `lamda^(3)+mu^(3)+5` is

Options

A. 676
B. 1130
C. 1348
D. 1077

Answer

C

Explanation

Let `D` be the mid-point of side BC of `DeltaABC`. The coordinate of D are `((lamda-1)/2,4,(mu+2)/2)`.
So direction of `vec(AD)` are proportional to `(lamda-5)/2,1,(mu-8)/2` and its direction cosines are
`(lamda-5)/(2k),1/k,(mu-8)/(2k)` where `k=sqrt(((lamda-5)/2)^(2)+1+((m8-u)/2)^(2))`

It is given that AD is equally inclined to the coordinates axes.
`:.(lamda-5)/(2k)=1/k=(mu-8)/(2k)=+-1/(sqrt(3))`
`implieslamda-5=2mu-8=2implieslamda=7,mu=10`
`:.lamda^(3)+5=7^(3)+10^(3)+5=1348`