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(D) The given system of equations is:$x - 2y + 5z = 0$  $(1)$$-2x + 4y + z = 0$  $(2)$$-7x + 14y + 9z = 0$  $(3)$First,we calculate the determinant of

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$8$

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(D) The given system of equations is:$x - 2y + 5z = 0$  $(1)$$-2x + 4y + z = 0$  $(2)$$-7x + 14y + 9z = 0$  $(3)$First,we calculate the determinant of the coefficient matrix:$\Delta = \begin{vmatrix} 1 & -2 & 5 \ -2 & 4 & 1 \ -7 & 14 & 9 \end{vmatrix} = 1(36 - 14) - (-2)(-18 + 7) + 5(-28 + 28) = 1(22) + 2(-11) + 0 = 22 - 22 = 0$.Since $\Delta = 0$,the system has infinitely many solutions.From $(1)$ and $(2)$,we have:$x - 2y = -5z$$-2x + 4y = -z$Multiplying the first equation by $2$,we get $2x - 4y = -10z$. Adding this to the second equation gives $0 = -11z$,so $z = 0$.Substituting $z = 0$ into $(1)$,we get $x - 2y = 0$,which implies $x = 2y$.Let $y = k$,where $k$ is an integer. Then $x = 2k$ and $z = 0$.The condition $15 \leq x^{2} + y^{2} + z^{2} \leq 150$ becomes:$15 \leq (2k)^{2} + k^{2} + 0^{2} \leq 150$$15 \leq 5k^{2} \leq 150$$3 \leq k^{2} \leq 30$.Since $k$ is an integer,$k^{2}$ can be $4, 9, 16, 25$.Thus,$k \in \{ \pm 2, \pm 3, \pm 4, \pm 5 \}$.There are $8$ possible values for $k$,each corresponding to a unique solution $(x, y, z)$.Therefore,the number of elements in $S$ is $8$.

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

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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