Question

There are three points $(a,x),(b,y)$ and $(c,z)$ such that straight lines joining any two of them are not equally inclined to the coordinate axes where $a,b,c,x,y,z\in R$.
If $\left|\begin{array}{ccc}x+a& y+b& z+c\\ y+b& z+c& x+a\\ z+c& x+a& y+b\end{array}\right|=0$ and $a+c=-b$, then $x,-\frac{y}{2},z$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

The correct option is A A.P.

$△=\left|\begin{array}{ccc}x+a& y+b& z+c\\ y+b& z+c& x+a\\ z+c& x+a& y+b\end{array}\right|=0$

$C_1\to C_1+C_2+C_3$

$△=\left|\begin{array}{ccc}x+a+y+b+z+c& y+b& z+c\\ x+a+y+b+z+c& z+c& x+a\\ x+a+y+b+z+c& x+a& y+b\end{array}\right|=0$

$C_1\to \frac{C_1}{x+a+y+b+z+c}$

$△=(x+a+y+b+z+c)\left|\begin{array}{ccc}1& y+b& z+c\\ 1& z+c& x+a\\ 1& x+a& y+b\end{array}\right|=0$

$R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

$△=(x+a+y+b+z+c)\left|\begin{array}{ccc}0& (y+b)-(z+c)& (z+c)-(x+a)\\ 0& (z+c)-(x+a)& (x+a)-(y+b)\\ 1& x+a& y+b\end{array}\right|=0$

$△=(x+y+z+a+b+c)[\left\{((y+b)-(z+c))((x+a)-(y+b))\right\}+\left\{((z+c)-(x+a))((x+a)-(z+c))\right\}]$

$=(x+a+y+b+z+c)[(y+b)(x+a)-{(y+b)}^2-(z+c)(x+a)+(z+c)(y+b)+(z+c)(x+a)-{(z+c)}^2-{(x+a)}^2+(x+a)(z+c)]$

$=(x+a+y+b+z+c)[-{(x+a)}^2-{(y+b)}^2-{(z+c)}^2+(x+a)(y+b)+(y+b)(z+c)+(z+c)(x+a)]$ the same terms of the opposite sign gets cancelled

Multiply by $-2$ and divide the above expression by $2$ we get

$=\frac{-(x+y+z)}{2}[2{(x+a)}^2+2{(y+b)}^2+2{(z+c)}^2-2(x+a)(y+b)-2(y+b)(z+c)-2(z+c)(x+a)]$ since $a+b+c=0$

$=\frac{-(x+y+z)}{2}[{(x+a)}^2+{(x+a)}^2+{(y+b)}^2+{(y+b)}^2+{(z+c)}^2+{(z+c)}^2-2(x+a)(y+b)-2(y+b)(z+c)-2(z+c)(x+a)]$ by splitting the terms

$=\frac{-(x+y+z)}{2}\frac{-(x+a+y+b+z+c)}{2}[{(x+a)}^2-2(x+a)(y+b)+{(x+a)}^2+{(y+b)}^2-2(y+b)(z+c)+{(y+b)}^2-2(z+c)(x+a)+{(z+c)}^2+{(z+c)}^2]$ by re-arranging

$=\frac{-(x+y+z)}{2}[{(x+a-y-b)}^2+{(y+b-z-c)}^2+{(z+c-x-a)}^2]$ using ${(a-b)}^2=a^2-2ab+b^2$

$\Rightarrow x+y+z=0$ or $x+a-y+b-z+c=0$ is impossible because the segment joining any two points is not equally on the axes.

$\Rightarrow x+z=-y$

$\Rightarrow x+z=2\left(\frac{-y}{2}\right)$

$\therefore x,\frac{-y}{2},z$ are in A.P