79252 If \(3 x+2 y+z=0, x+4 y+z=0,2 x+y+4 z=0\)
be a system of homogeneous equation, then

79253 The system of equations \(2 x+y-5=0\), \(x-2 y+1=9,2 x-14 y-a=0\), is consistent. Then, \(a\) is equal to

79254 The value of \(x\) obtained from the equation
\(\left|\begin{array}{ccc} \mathbf{x}+\boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\gamma} \\ \boldsymbol{\gamma} & \mathbf{x}+\boldsymbol{\beta} & \boldsymbol{\alpha} \\ \boldsymbol{\alpha} & \boldsymbol{\beta} & \mathbf{x}+\boldsymbol{\gamma} \end{array}\right|=\mathbf{0} \text { will be }\)

79255 Find the value of \(k\) for which the simultaneous equations \(x+y+z=3 ; x+2 y+3 z=4\) and \(x+\) \(4 y+k z=6\) will not have a unique solution.

79252 If \(3 x+2 y+z=0, x+4 y+z=0,2 x+y+4 z=0\)
be a system of homogeneous equation, then

79253 The system of equations \(2 x+y-5=0\), \(x-2 y+1=9,2 x-14 y-a=0\), is consistent. Then, \(a\) is equal to

79254 The value of \(x\) obtained from the equation
\(\left|\begin{array}{ccc} \mathbf{x}+\boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\gamma} \\ \boldsymbol{\gamma} & \mathbf{x}+\boldsymbol{\beta} & \boldsymbol{\alpha} \\ \boldsymbol{\alpha} & \boldsymbol{\beta} & \mathbf{x}+\boldsymbol{\gamma} \end{array}\right|=\mathbf{0} \text { will be }\)

79255 Find the value of \(k\) for which the simultaneous equations \(x+y+z=3 ; x+2 y+3 z=4\) and \(x+\) \(4 y+k z=6\) will not have a unique solution.

79252 If \(3 x+2 y+z=0, x+4 y+z=0,2 x+y+4 z=0\)
be a system of homogeneous equation, then

79253 The system of equations \(2 x+y-5=0\), \(x-2 y+1=9,2 x-14 y-a=0\), is consistent. Then, \(a\) is equal to

79254 The value of \(x\) obtained from the equation
\(\left|\begin{array}{ccc} \mathbf{x}+\boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\gamma} \\ \boldsymbol{\gamma} & \mathbf{x}+\boldsymbol{\beta} & \boldsymbol{\alpha} \\ \boldsymbol{\alpha} & \boldsymbol{\beta} & \mathbf{x}+\boldsymbol{\gamma} \end{array}\right|=\mathbf{0} \text { will be }\)

79255 Find the value of \(k\) for which the simultaneous equations \(x+y+z=3 ; x+2 y+3 z=4\) and \(x+\) \(4 y+k z=6\) will not have a unique solution.

79252 If \(3 x+2 y+z=0, x+4 y+z=0,2 x+y+4 z=0\)
be a system of homogeneous equation, then

79253 The system of equations \(2 x+y-5=0\), \(x-2 y+1=9,2 x-14 y-a=0\), is consistent. Then, \(a\) is equal to

79254 The value of \(x\) obtained from the equation
\(\left|\begin{array}{ccc} \mathbf{x}+\boldsymbol{\alpha} & \boldsymbol{\beta} & \boldsymbol{\gamma} \\ \boldsymbol{\gamma} & \mathbf{x}+\boldsymbol{\beta} & \boldsymbol{\alpha} \\ \boldsymbol{\alpha} & \boldsymbol{\beta} & \mathbf{x}+\boldsymbol{\gamma} \end{array}\right|=\mathbf{0} \text { will be }\)

79255 Find the value of \(k\) for which the simultaneous equations \(x+y+z=3 ; x+2 y+3 z=4\) and \(x+\) \(4 y+k z=6\) will not have a unique solution.