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(D) Given the system of equations:$1) 3x - y - z = 0$$2) -3x + z = 0$$3) -3x + 2y + z = 0$From equation $(2)$,we get $z = 3x$.Substituting $z = 3x$ in

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$7$

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(D) Given the system of equations:$1) 3x - y - z = 0$$2) -3x + z = 0$$3) -3x + 2y + z = 0$From equation $(2)$,we get $z = 3x$.Substituting $z = 3x$ into equation $(1)$:$3x - y - 3x = 0 \Rightarrow y = 0$.Checking with equation $(3)$:$-3x + 2(0) + 3x = 0$,which is $0 = 0$. This is consistent.Thus,any point $(x, y, z)$ satisfying the system is of the form $(a, 0, 3a)$ where $a$ is an integer.We are given the condition $x^2 + y^2 + z^2 \leq 100$.Substituting the point $(a, 0, 3a)$:$a^2 + 0^2 + (3a)^2 \leq 100$$a^2 + 9a^2 \leq 100$$10a^2 \leq 100$$a^2 \leq 10$Since $a$ is an integer,the possible values for $a$ are $a \in \{-3, -2, -1, 0, 1, 2, 3\}$.Counting these values,we get $7$ possible points.

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:

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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: