If the system of equations $2x + 9y + 5z = 8$, $2x + 3y - z = -4$, $x - 2z = -5$ has an infinite number of solutions $x = -5 + at$, $y = 2 + bt$, $z = ct$, $t \in R$, then $a$, $b$, $c$ respectively are

If $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix}$,then $2x - y + z = $

If $x=\alpha, y=\beta, z=\gamma$ is the unique solution of the system of linear equations $2x-3y+5z=12$, $5x+2y+3z=11$, and $x+2y-3z=-3$, then $2\alpha+5\beta+3\gamma=$

Let the system of equations: $2x + 3y + 5z = 9$,$7x + 3y - 2z = 8$,$12x + 3y - (4 + \lambda)z = 16 - \mu$ have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4x = 3y$ is

Solve the system of equations $2x + 5y = 1$ and $3x + 2y = 7$.

In solving a system of linear equations $AX=B$ by Cramer's rule,in the usual notation,if $\Delta_1=\left|\begin{array}{ccc}-11 & 1 & -7 \\ -4 & 1 & -2 \\ 5 & 1 & 1\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ccc}4 & 1 & -11 \\ 1 & 1 & -4 \\ 4 & 1 & 5\end{array}\right|$,then $X=$