JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 7)

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 $$ \le $$ x2 + y2 + z2 $$ \le $$ 150. Then, the number of elements in the set S is equal to ______ .
Answer
8

Explanation

$$x - 2y + 5z = 0$$ ....(1)

$$ - 2x + 4y + z = 0$$ .....(2)

$$ - 7x + 14y + 9z = 0$$ ....(3)

2.(1) + (2) we get z = 0, x = 2y

15 $$ \le $$ 4y2 + y2 $$ \le $$ 150

$$ \Rightarrow $$ 3 $$ \le $$ y2 $$ \le $$ 30

$$y \in \left[ { - \sqrt {30} , - \sqrt 3 } \right] \cup \left[ {\sqrt 3 ,\sqrt {30} } \right]$$

$$y = \pm 2,\, \pm 3,\, \pm 4,\, \pm 5$$

$$ \therefore $$ no. of integer's in S is 8

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