The number of real values of $\alpha$ for which the system of equations
$x+3y+5z=\alpha x$
$5x+y+3z=\alpha y$
$3x+5y+z=\alpha z$
has infinite number of solutions is

Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{*}, y^{*}, z^{*})$. If $(\alpha, x^{*}), (y^{*}, \alpha)$ and $(x^{*}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$

The system of equations $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = -18$ has

Let $a, b, c$ be positive real numbers. The following system of equations in $x, y, z$:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$
$\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
$-\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
has:

Let $S$ be the set of all integer solutions,$(x, y, z)$,of the system of equations
$x-2y+5z=0$
$-2x+4y+z=0$
$-7x+14y+9z=0$
such that $15 \leq x^{2}+y^{2}+z^{2} \leq 150$. Then,the number of elements in the set $S$ is equal to