Q27 / 75 · JEE Main 2025 (24 Jan Shift 2) · Mathematics

If the system of equations $\begin{aligned} & x+2 y-3 z=2 \\ & 2 x+\lambda y+5 z=5 \\ & 14 x+3 y+\mu z=33 \end{aligned}$ has infinitely many solutions, then $\lambda+\mu$ is equal to :

  • A
    13
  • B
    10
  • Correct answer
    C
    12
  • D
    11

Solution

$\begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{array}\right|=0 \Rightarrow \lambda \mu+42 \lambda-4 \mu+107=0 \\ & \Delta_1=\left|\begin{array}{ccc} 2 & 2 & -3 \\ 5 & \lambda & 5 \\ 33 & 3 & \mu \end{array}\right|=0 \Rightarrow 2 \lambda \mu+99 \lambda-10 \mu+255=0 \\ & \Delta_2=\left|\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & \mu \end{array}\right|=0 \Rightarrow \mu=13 \end{aligned}$ Also, $\lambda=-1$ Hence, $\lambda+\mu=13-1=12$

Hence, the correct answer is option: 'C'.

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