Question

Let $S$ be the set of all integer solutions, $(x,y,z)$, of the system of equations

$$\begin{gathered} x-2y+5z=0 \\ -2x+4y+z=0 \\ -7x+14y+9z=0 \end{gathered}$$

such that $15\le x^2+y^2+z^2\le 150$.

Then, the number of elements in the set $S$ is equal to:

Answer 1

Determining the number of elements in $S$

We have the system of equations as:

$$\begin{gathered} x-2y+5z=0.....\left(i\right) \\ -2x+4y+z=0.....\left(ii\right) \\ -7x+14y+9z=0.....\left(iv\right) \end{gathered}$$

such that $15\le x^2+y^2+z^2\le 150$

Taking the determinant of coefficient of the system of equations

$∆=\left|\begin{array}{ccc}1& -2& 5\\ -2& 4& 1\\ -7& 14& 9\end{array}\right|$

Considering $x=k$ and substituting it into the equation $\left(i\right)\&\left(ii\right)$

$$\begin{gathered} \Rightarrow k-2y+5z=0 \\ \Rightarrow -2k+4y+z=0 \end{gathered}$$

Solving it we have $z=0,y=\frac{k}{2}$

$\therefore x,y,z$are integer

$\Rightarrow$$k$ is even an integer.

Finding the values of $k$

The given condition $15\le x^2+y^2+z^2\le 150$

Putting $x=k,y=\frac{k}{2},z=0$ in the above condition

$$\begin{gathered} \Rightarrow 15\le k^2+{\left(\frac{k}{2}\right)}^2+0\le 150 \\ \Rightarrow 12\le k^2\le 120 \\ \Rightarrow k=\pm 4,\pm 6,\pm 8,\pm 10 \end{gathered}$$

Since the number of elements of $S$ is equal to the number of values of $k$

Thus, the number of elements of $S$ $=8$