If the system of linear equations given by $x+y+z=3$,$2x+2y-z=3$,and $x+y-z=1$ is consistent and if $(x_0, y_0, z_0)$ is a solution,then $2x_0+2y_0+z_0=$

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations
$2x - y + 2z = 2$
$x - 2y + \lambda z = -4$
$x + \lambda y + z = 4$
has no solution. Then the set $S$

The sum of three numbers is $6$. Thrice the third number when added to the first number gives $7$. On adding three times the first number to the sum of the second and third numbers,we get $12$. The product of these numbers is:

If the system of linear equations $x - 2y + kz = 1$,$2x + y + z = 2$,and $3x - y - kz = 3$ has a non-zero solution $(x, y, z) \neq 0$,then $(x, y)$ lies on the straight line whose equation is

Consider the system of linear equations in $x, y, z$: $x+2y+tz=0, 6x+y+5tz=0, 3x+t^2y+z=0$. If this system has infinitely many solutions for all $t \in R$,then the determinant of the coefficient matrix must be zero for all $t$. Let $D(t)$ be the determinant of the coefficient matrix. If $D(t) = 0$ for all $t$,analyze the condition.

If the system of simultaneous linear equations $3x - 4y + kz + 13 = 0$,$x + 2y - z - 9 = 0$,and $kx - y + 3z + 7 = 0$ has a unique solution $x = \alpha, y = \beta, z = \gamma$ for $k \neq m$ and $2\beta - \gamma = 8$,then $\alpha + m =$