# 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 ______. Option: 1 6 Option: 2 4 Option: 3 10 Option: 4 8

The system of equations are

$x-2y+5z=0$

$-2x+4y+z=0$

$-7x+14y+9z=0$

$\begin{array}{l} \Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ -2 & 4 & 1 \\ -7 & 14 & 9 \end{array}\right|=0 \\ \text {Let } x=k \end{array}$

\begin{aligned} \Rightarrow & \text { Put in }(1) \&(2) \\ & \mathrm{k}-2 \mathrm{y}+5 \mathrm{z}=0 \\ &-2 \mathrm{k}+4 \mathrm{y}+\mathrm{z}=0 \\ & \mathrm{z}=0, \mathrm{y}=\frac{\mathrm{k}}{2} \\ \therefore & \mathrm{x}, \mathrm{y}, \mathrm{z} \text { are integer } \end{aligned} \\\Rightarrow \mathrm{k} \text{ is even integer}

\\\text{Now} x=k, y=\frac{k}{2}, z=0\\\text{ put in condition }\\15 \leq \mathrm{k}^{2}+\left(\frac{\mathrm{k}}{2}\right)^{2}+0 \leq 150\\ \begin{aligned} & 12 \leq \mathrm{k}^{2} \leq 120 \\ \Rightarrow & \mathrm{k}=\pm 4,\pm 6,\pm 8,\pm 10 \end{aligned}\\\Rightarrow Number of element in \mathrm{S}=8

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