The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

The set of equations $x - y + 3z = 2$,$2x - y + z = 4$,and $x - 2y + \alpha z = 3$ has:

$A$ manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below:

Market	$x, y, z$
$I$	$10,000, 2,000, 18,000$
$II$	$6,000, 20,000, 8,000$

If unit sale prices of $x, y$ and $z$ are Rs. $2.50$,Rs. $1.50$ and Rs. $1.00$ respectively,find the total revenue in each market with the help of matrix algebra.

The values of $a$ and $b$,for which the system of equations $2x + 3y + 6z = 8$,$x + 2y + az = 5$,and $3x + 5y + 9z = b$ has no solution,are:

Let $(x, y, z)$ be points with integer coordinates satisfying the system of homogeneous equations:
$3x - y - z = 0$,$-3x + z = 0$,$-3x + 2y + z = 0$.
Then the number of such points for which $x^2 + y^2 + z^2 \leq 100$ is:

If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .