Question

Consider the following linear equations $ax+by+cz=0$, $bx+cy+az=0$, $cx+ay+bz=0$.
Match the elements of List I with elements of List II.

Answer 1

$\left|A\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=-\frac{1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$
A) For $a+b+c\ne 0$
$\Rightarrow [{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=0$
$\Rightarrow a=b=c\ne 0$
And the given equation represents three identical planes $x+y+z=0$
B) For $\left|A\right|=0$
$\Rightarrow$ the solution of equation is not unique
$ax+by+cz=0\Rightarrow by+cz=(b+c)x$ and $cy+az=(c+a)x$
$\Rightarrow (b+c)y+(c+a)z=(b+c)x+(c+a)x\Rightarrow x=y=z$
C) $\left|A\right|\ne 0$, equation have a unique solution $x=y=z=0$ and three planes meet at a point.
D) $\Rightarrow a=b=c=0$; $x,y,z$ can take any values and hence the given equation represent the whole of three dimensional plane.