`lambda x + 2y + 2z = 5, 2lambda x + 3y + 5z = 8, 4x + lambda y + 6z = 10` for the system of equation check the correct option.

The system of linear equations lambdax+2y+2z=5 2lambda x+3y+5z=8 4x+lambday+6z=10 has :

Given 2x+4y+z=1,lambda x+2y+z=2,x+y-lambda z=3 then one of the value of a such that the given system of equations has no solution,is

Given 2 x-y+2 z=1 , x-2 y+z=-4 , and x+y+lambda z =4 . Then the. value of lambda such that the given system of equation has no solution is

Let x+y+z=6,4x+lambda y-lambda z=0,3x+2y-4z=-5 The value of lambda for which given system of equations does not have a unique solution is

Given , 2x-y+2z=2, x-2y+z=-4, x+y+ lambda z=4, then the value of lambda such that the given system of equations has no solution is :

Given , 2x-y+2z=2, x-2y+z=-4, x+y+ lambda z=4, then the value of lambda such that the given system of equations has no solution is :

If the system of linear equations x + y + z = 5 x + 2y + 2z = 6 x + 3y + lambda z = mu, (lambda, mu in R) has infinitely many solutions, then the value of lambda + mu is :

Show that the system of eqations 3x - y + 4z = 3, x + 2y - 3z = -2, 6x + 5y + lambda z = -3 has atleast one solution for every real lambda , Find the set of solutions when lambda = -5