If the homogeneous system of linear equations $x-2y+3z=0, 2x+4y-5z=0, 3x+\lambda y+\mu z=0$ has a non-trivial solution, then $8\mu+11\lambda=$

If the solution of the system of simultaneous linear equations $x+y-z=6$,$3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha+\beta=$

Let $S$ be the set of all integer solutions,$(x, y, z)$,of the system of equations
$x-2y+5z=0$
$-2x+4y+z=0$
$-7x+14y+9z=0$
such that $15 \leq x^{2}+y^{2}+z^{2} \leq 150$. Then,the number of elements in the set $S$ is equal to

If the system of equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method, then:

The values of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6$,$3x+5y+5z=26$,and $x+2y+\lambda z=\mu$ has no solution are:

If the values $x=\alpha, y=\beta, z=\gamma$ satisfy all the $3$ equations $x+2y+3z=4$,$3x+y+z=3$ and $x+3y+3z=2$,then $3\alpha+\gamma=$