Question

Consider the following system of equations :
$ax+by+cz=0az+bx+cy=0ay+bz+cx=0$
$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ \text{(I)}& \text{If}a+b+c\ne 0\text{and}& \text{(P)}& \text{Planes meet only at one point}\\ & (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0.& & \\ \text{(II)}& \text{If}a+b+c=0\text{and}& \left(Q\right)& \text{Equations represent the line}x=y=z\\ & (a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ne 0& & \\ \text{(III)}& \text{If}a+b+c\ne 0\text{and}& \text{(R)}& \text{Equations represent identical planes}\\ & (a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ne 0& & \\ \text{(IV)}& \text{If}a+b+c=0\text{and}& \text{(S)}& \text{The solution of the system represents}\\ & (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0& & \text{whole of the three dimensional space}\end{array}$


Which of the following is the "INCORRECT" option?

• A. $\left(IV\right)\to \left(S\right)$
• B. $\left(II\right)\to \left(Q\right)$
• C. $\left(I\right)\to \left(S\right)$
• D. $\left(III\right)\to \left(P\right)$

Answer 1

The correct option is C $\left(I\right)\to \left(S\right)$

$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|\Rightarrow \Delta =a(cb-a^2)-b(b^2-ac)+c(ab-c^2)\Rightarrow \Delta =3abc-a^3-b^3-c^3\Rightarrow \Delta =-\frac{1}{2}(a+b+c)[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

$\left(I\right)$
$a+b+c\ne 0\text{and}(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0\Rightarrow a=b=c$
So, the system of equations represents identical planes.
$\left(I\right)\to \left(R\right)$

$\left(II\right)$
$a+b+c=0\text{and}(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ne 0$
$c=-(a+b)$
The system of equations becomes
$ax+by=(a+b)zaz+bx=(a+b)yay+bz=(a+b)x$
This represents the line $x=y=z$
$\left(II\right)\to \left(Q\right)$

$\left(III\right)$
$a+b+c\ne 0\text{and}(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ne 0$
$\Rightarrow \Delta \ne 0$
So, there is only trivial solution, $(0,0,0)$
Hence, planes intersect only at a point.
$\left(III\right)\to \left(P\right)$

$\left(IV\right)$
$a+b+c=0\text{and}(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$
$\Rightarrow a=b=c=0$
$\therefore$ The solution of the system is $R^3$,
i.e., whole three dimensional space.
$\left(IV\right)\to \left(S\right)$