`A B C` is a triangle where `A=(2,3,5)B=(-1,2,2)a n dC(lambda,5mu)` , if the median through `A` is equally inclined to the axes then: `lambda=mu=5` (b) `lambda=5,mu=7` `lambda=6,mu=9` (d) `lambda=0,mu=0`

ABC is a triangle and A=(2,3,5),B=(-1,3,2) and C=(lambda,5,mu). If the median through A is equally inclined to the axes,then find the value of lambda and mu

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

A(2,3,5),B(-1,3,2),C( lambda ,5, mu ) are the vertices of a triangle.If the median through A is equally inclined to co-ordinate axes then the value of 3 lambda-2 mu is

Let A(2, 3, 5), B(-1, 3, 2), C(lambda, 5, mu) are the vertices of a triangle and its median through A(I.e.,) AD is equally inclined to the coordinates axes. Q. Projection of AB onBC is

Let A(2, 3, 5), B(-1, 3, 2), C(lambda, 5, mu) are the vertices of a triangle and its median through A(I.e.,) AD is equally inclined to the coordinates axes. Q. On the basis of the above information answer the following Q. The value of 2lambda-mu is equal to

Find the cartesian form of the equation of the plane. vecr=(lambda-mu)hati+(1-mu)hatj+(2lambda+3mu)hatk

If y=2 is the directrix and (0,1) is the vertex of the parabola x^2+lambday+mu=0 , then (a)lambda=4 (b) mu=8 (c) lambda=-8 (d) mu=4

The system of equations: x+y+z=5x+2y+3z=9x+3y+lambda z=mu has a unique solution,if lambda=5,mu=13(b)lambda!=5lambda=5,mu!=13(d)mu!=13

If the system of linear equation x+y+z=6,x+2y+3c=14, and 2x+5y+lambda z=mu(lambda,mu epsilon R) has a unique solution,then lambda=8 b.lambda=8,mu=36c.lambda=8,mu!=36