If $\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$,then the value of $x^2 + y^2 + z^2 =$

For what value of $k$ does the following system of equations possess a non-trivial solution?
$x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 3y - 4z = 0$

If the system of linear equations $2x + 2y + 3z = a$,$3x - y + 5z = b$,and $x - 3y + 2z = c$,where $a, b, c$ are non-zero real numbers,has more than one solution,then:

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 equations $x + y - z = 0, 3x - \alpha y - 3z = 0, x - 3y + z = 0$ has a non-zero solution,then $\alpha = $

For the system of linear equations $a x+y+z=1$,$x+a y+z=1$,$x+y+a z=\beta$,which one of the following statements is $NOT$ correct?