A solution set of the equations `x+2y+z=1`, `x+3y+4z=k`, `x+5y+10z=k^(2)` is

A solutions set of the equations x+2y+z=1,x+3y+4z=k,x+5y+10z= k^2 (5 lambda-1,1-3 lambda,lambda) (1+5 lambda,-3 lambda,lambda) (1+6 lambda,-2 lambda,lambda) (1-6"lambda,lambda,lambda)

A solution set of the equations x-3y-8z+10=0,3x+y-4z=0,2x+5y+6z=13,is-

If A and B are the two real values of k for which the system of equations x+2y+z=1, x+3y+4z=k, x+5y+10z=k^(2) is consistent, then A+B =

A solution set of the equations x-3y-8z+10=0,3x+y-4z=0, 2x+5y+6z=13 is (A)(2K+1,3-2K,K) (B) (2K-1,3+2K,K) (C) (1-2K,3-2K,K) (D) (2K-1,3-2K,K)

For the equations x+2y+3z=1, 2x+y+3z=2, 5x+5y+9z=4

The equations x+2y+3z=1 2x+y+3z=1 5x+5y+9z=4

If (x,y,z) is the solution of the equations x - y - 2z = 3 2x + y + 4z = 5 4x - y - 2z = 11 then the value ofy equals

The number of distinct real values of K such that the system of equations x+2y+z=1 , x+3y+4z=K, x+5y+10z = K^2 has infinitely many solutions is :

If (x,y,z) is the solution of the equations 4x + y = 7 3y + 4z = 5 5x + 3z = 2 Then the value of x + y + z equals

The system of equations kx + y + z =1, x + ky + z = k and x + y + zk = k^(2) has no solution if k is equal to :