If the set of equations $x+2y+3z=6$,$x+3y+5z=9$,and $2x+5y+az=b$ has a unique solution,then:

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?

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.

The solution of equations $x + y = 10$,$2x + y = 18$,and $4x - 3y = 26$ is:

If the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$,and $x + y + 3z = 4$ has infinitely many solutions,then $p=$

If the system of equations
$x-2y+3z=9$
$2x+y+z=b$
$x-7y+az=24$
has infinitely many solutions,then $a-b$ is equal to