Q38 / 57 · JEE Advanced 2009 (Paper 2) · Mathematics
Let $(x, y, z)$ be points with integer coordinates satisfying the system of homogeneous equations $3 x-y-z=0,-3 x+z=0,-3 x+2 y+z=0$. Then, the number of such points for which $x^2+y^2+z^2 \leq 100$ is
Correct answer: 7
Solution
Given, $\quad 3 x-y-z=0$
and
$$
\begin{array}{r}
-3 x+2 y+z=0 \\
-3 x+z=0
\end{array}
$$
On adding Eqs. (i) and (ii), we get $y=0$
So, $\quad 3 x=z$
Now, $\quad x^2+y^2+z^2 \leq 100$
$$
\begin{array}{lc}
\Rightarrow & x^2+(3 x)^2+0 \leq 100 \\
\Rightarrow & 10 x^2 \leq 100 \\
\Rightarrow & x^2 \leq 10 \\
\therefore & x=-3,-2,-1,0,1,2,3
\end{array}
$$
So, number of such 7 points are possible.
Hence, the correct answer is 7.