JEE Main 2026 — 5 April, Evening Shift — Question 28

Let $\mathrm{M}=\left[\begin{array}{lll}\mathrm{a}_{1} & \mathrm{a}_{2} & \mathrm{a}_{3}\\ \mathrm{~b}_{1} & \mathrm{~b}_{2} & \mathrm{~b}_{3}\\ \mathrm{c}_{1} & \mathrm{c}_{2} & \mathrm{c}_{3}\end{array}\right]$ then $\mathrm{M}\left(\begin{array}{l}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right) \Rightarrow \mathrm{a}_{1}=1, \mathrm{~b}_{1}=2, \mathrm{c}_{1}=3$ $\mathrm{M}\left(\begin{array}{l}0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}0\\ 1\\ 2\end{array}\right) \Rightarrow \mathrm{a}_{2}=0, \mathrm{~b}_{2}=1, \mathrm{c}_{2}=2$ $\& \mathrm{M}\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 1\\ 1\end{array}\right) \Rightarrow \mathrm{a}_{3}=-1, \mathrm{~b}_{3}=1, \mathrm{c}_{3}=1$ Now $M\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}1\\ 7\\ 11\end{array}\right)$ $\Rightarrow \mathrm{x}+0 \mathrm{y}-\mathrm{z}=1$

$$\begin{aligned} & 2 x+y+z=7 \end{aligned}$$

$3 x+2 y+z=11$

on solving $\mathrm{x}=2, \mathrm{y}=2, \mathrm{z}=1$ $\therefore \mathrm{x}+\mathrm{y}+\mathrm{z}=5$