$3 x-y-z $$ =0 $, $-3 x+z $$ =0 $, $-3 x+2 y+z $$ =0 .$
Then the number of such points for which $x^2+y^2+z^2 \leq 100$ is
- A$3$
- B$9$
- C$5$
- ✓$7$
$3 x-y-z $$ =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
and substituting $y=0$ in third equation, we get, $z=3 x$
So any point which satisfies given system can be taken as, $(a, o, 3 a )$
Now for this point to lie inside inside a sphere of radius $10$ centered at origin.
$\Rightarrow a ^2+ o ^2+(3 a )^2 < 10^2$
$\Rightarrow a ^2 < 10$
So, possible integral values of a are $-3,-2,-1,0,1,2,3$
Hence, number of such points is $7$ .
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$( S 1):|(\overrightarrow{ a } \times \overrightarrow{ b })+(\overrightarrow{ c } \times \overrightarrow{ b })|-|\overrightarrow{ c }|=6(2 \sqrt{2}-1)$
$( S 2): \angle ABC =\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then