JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Determinants

Infinitely many solutions require the coefficient determinant to vanish. Expanding $\left|\begin{array}{ccc}2& 3& 5\\ 7& 3& -2\\ 12& 3& -(4+\lambda )\end{array}\right|=0$ reduces to $15\lambda -75=0$, so $\lambda =5$. Subtracting the first equation from the second gives $5x-7z=-1$, and subtracting the second from the third gives $5x-7z=8-\mu$; for these to agree, $8-\mu =-1$, so $\mu =9$. The circle is centred at $(5,9)$ and touches the line $4x-3y=0$, so its radius equals the perpendicular distance $\frac{|4(5)-3(9)|}{\sqrt{4^2+3^2}}=\frac{7}{5}$. The required radius is $\frac{7}{5}$.