JEE Main 2025 Apr 7 Shift 1, Mathematics Q17: Linear Equations And Cramer’s Rule

Step 1: Find $\lambda$ (coefficient determinant = 0).

$\left|\begin{array}{c}235\\ 73-2\\ 123-(4+\lambda )\end{array}\right|=0$

Expand along row 1:

$2[3\cdot (-(4+\lambda ))-(-2)(3)]-3[7\cdot (-(4+\lambda ))-(-2)(12)]+5[7(3)-3(12)]=0$

$=2[-3(4+\lambda )+6]-3[-7(4+\lambda )+24]+5[21-36]$

$=2(-12-3\lambda +6)-3(-28-7\lambda +24)+5(-15)$

$=2(-6-3\lambda )-3(-4-7\lambda )-75$

$=-12-6\lambda +12+21\lambda -75=15\lambda -75$

Setting equal to zero: $\boxed{\lambda =5}$.

Step 2: Find $\mu$ (any augmented-column determinant also equals 0).

$\begin{vmatrix} 9 & 3 & 5\\ 8 & 3 & -2\\ 16-\mu & 3 & -9 \end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} 1 & 0 & 7\\ \mu-8 & 0 & 7\\ 16-\mu & 3 & -9 \end{vmatrix}=0$

$\Rightarrow 7-7(\mu -8)=0$

$\Rightarrow 1-(\mu -8)=0$

$\therefore \boxed{\mu =9}$

Step 3: Distance from $(5,9)$ to the line $4x-3y=0$.

$r=\frac{|4(5)-3(9)|}{\sqrt{4^2+3^2}}=\frac{|20-27|}{5}=\frac{7}{5}$

$\therefore$ Option (2).