JEE Main 2025 Jan 29 Shift 2, Mathematics Q24: Properties Of Determinant

Simplify the determinant. Add the second and third rows to the first row. Using $1+\omega +{\omega }^2=0$, every entry of the new first row becomes $z$.

Factor $z$ out of row 1 and then perform column operations $C_2\to C_2-\omega C_1$, $C_3\to C_3-{\omega }^2{C}_1$, which reduces the lower-right $2\times 2$ block to a diagonal form. Expanding gives $z\cdot z^2=z^3=1$.

Hence $z\in \{1,\omega ,{\omega }^2\}$.

Apply $|z-a|=|z+b|$ for each value of $z$.

(i) $z=1$: $|1-a|=|1+b|$, giving $a-b=2$ (the other sign gives $a+b=0$, excluded). Pairs with $b\in \{-3,-2,0,1\}$ (skip $b=-1$ which forces $a+b=0$): $\{(-1,-3),(0,-2),(2,0),(3,1)\}$. 4 pairs.

(ii) $z=\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$: equate squared moduli ${\left(a+\frac{1}{2}\right)}^2={\left(b-\frac{1}{2}\right)}^2$, giving $a-b=-1$ (the other root gives $a+b=0$). Integer solutions with $a,b\in [-3,3]$: $\{(-3,-2),(-2,-1),(-1,0),(0,1),(1,2),(2,3)\}$. 6 pairs.

(iii) $z={\omega }^2$: yields the same equation as $z=\omega$, no new pairs.

The two sets are disjoint, so total $=4+6=10$.