QJEEM22Madf7e: JEE Main 2022 Mathematics - Class 12 JEEM22Madf7e Solution: Let the system of linear equations x + y + α z = 2 3x + y + z = …

Given system of equations

$$x+y+az=2$$ ..... (i)

$$3x+y+z=4$$ ..... (ii)

$$x+2z=1$$ ..... (iii)

Solving (i), (ii) and (iii), we get

x = 1, y = 1, z = 0 (and for unique solution a $$\ne$$ $$-$$3)

Now, ($$\alpha$$, 1), (1, $$\alpha$$) and (1, $$-$$1) are collinear

$$\therefore$$ $$\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& -1& 1\end{array}\right|=0$$

$$\Rightarrow \alpha (\alpha +1)-1(0)+1(-1-\alpha )=0$$

$$\Rightarrow {\alpha }^2-1=0$$

$$\therefore$$ $$\alpha =\pm 1$$

$$\therefore$$ Sum of absolute values of $$\alpha =1+1=2$$