JEE Advanced · Mathematics · 19. Determinants
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
- A 5
- B 9
- C 1
- D 7
Answer & Solution
Correct Answer
(D) 7
Step-by-step Solution
Detailed explanation
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.
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.
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