JEE Main 2024 — 9 April, Shift 1 — Question 5

$3 \mathrm{x}+5 \mathrm{y}+\lambda \mathrm{z}=3$

$7 \mathrm{x}+11 \mathrm{y}-9 \mathrm{z}=2$

$97 \mathrm{x}+155 \mathrm{y}-189 \mathrm{z}=\mu$

$93 \mathrm{x}+155 \mathrm{y}+31 \lambda \mathrm{z}=93$

$97 x+155 y-189 z=\mu$

$$\frac{-\quad -\quad +\quad -}{-4x + (31\lambda + 189)z = 93 - \mu}$$ $$1085x + 1705y - 1395z = 310$$ $$1067x + 1705y - 2079z = 11\mu$$ $$\underline{-\quad -\quad +\quad -}$$

$18 \mathrm{x}+684 \mathrm{z}=310-11 \mu$

$-36 x+9(31 \lambda+189) z=9(93-\mu)$

$36 x+1368 z=2(310-11 \mu)$ $(279 \lambda+3069) \mathrm{z}=1457-31 \mu$

for infinite solutions -

$\lambda=\frac{-3069}{279}=\frac{-341}{31}$

$\mu=\frac{1457}{31}$

$\mu+2 \lambda=\frac{1457-682}{31}=\frac{775}{31}=25$