JEE Main 2024 — 1 February, Shift 2 — Question 17

$$\begin{cases} x+2y+3z=5,\\[4pt] 2x+3y+z=9,\\[4pt] 4x+3y+\lambda z=\mu. \end{cases}$$

For the system to have infinitely many solutions we must have

$$\Delta=\begin{vmatrix} 1 & 2 & 3\\[4pt] 2 & 3 & 1\\[4pt] 4 & 3 & \lambda \end{vmatrix}=0,\qquad \Delta_1=\begin{vmatrix} 5 & 2 & 3\\[4pt] 9 & 3 & 1\\[4pt] \mu & 3 & \lambda \end{vmatrix}=0,$$

(and similarly $\Delta_2=\Delta_3=0$, but $\Delta$ and one of the $\Delta_i$ suffice to determine $\lambda,\mu$.)

Compute $\Delta$:

$$\Delta =1(3\lambda-1\cdot 3)-2(2\lambda-1\cdot 4)+3(2\cdot 3-3\cdot 4)$$ $$= (3\lambda-3)-2(2\lambda-4)+3(6-12) =3\lambda-3-4\lambda+8+3(-6)$$ $$=-\lambda-13.$$

Thus $\Delta=0\Rightarrow -\lambda-13=0\Rightarrow \boxed{\lambda=-13}.$

Now compute $\Delta_1$ and substitute $\lambda=-13$:

$$\Delta_1 =5(3\lambda-1\cdot 3)-2(9\lambda-1\cdot\mu)+3(9\cdot 3-3\mu).$$

With $\lambda=-13$,

$$3\lambda-3=3(-13)-3=-39-3=-42,$$ $$9\lambda-\mu=9(-13)-\mu=-117-\mu,$$ $$9\cdot 3-3\mu=27-3\mu.$$

Therefore

$$\Delta_1=5(-42)-2(-117-\mu)+3(27-3\mu).$$

Simplify:

$$\Delta_1=-210+234+2\mu+81-9\mu =105-7\mu.$$

Setting $\Delta_1=0$ gives

$$105-7\mu=0\quad\Rightarrow\quad \boxed{\mu=15}.$$

Hence for $\lambda=-13,\;\mu=15$ the system has infinitely many solutions. Finally,

$$\lambda+2\mu = -13 + 2(15) = -13 + 30 = \boxed{17}.$$