Consider the following linear equations:ax+by | vedclass

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(D) The determinant of the system is $\Delta = \begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix} = -\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)

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$A-r, B-q, C-p, D-s$

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(D) The determinant of the system is $\Delta = \begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix} = -\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]$.$(A)$ If $a+b+c 
eq 0$ and $a^2+b^2+c^2 = ab+bc+ca$,then $a=b=c 
eq 0$. The equations become $a(x+y+z)=0$,which represents identical planes. Thus,$(A)-(r)$.$(B)$ If $a+b+c=0$ and $a^2+b^2+c^2 
eq ab+bc+ca$,then $\Delta=0$. The system has infinitely many solutions. Solving the equations leads to $x=y=z$. Thus,$(B)-(q)$.$(C)$ If $a+b+c 
eq 0$ and $a^2+b^2+c^2 
eq ab+bc+ca$,then $\Delta 
eq 0$. The system has a unique solution $(0,0,0)$,representing planes meeting at a single point. Thus,$(C)-(p)$.$(D)$ If $a+b+c=0$ and $a^2+b^2+c^2 = ab+bc+ca$,then $a=b=c=0$. The equations become $0=0$,representing the whole three-dimensional space. Thus,$(D)-(s)$.

Column $I$	Column $II$
$(A)$ $a+b+c \neq 0$ and $a^2+b^2+c^2=ab+bc+ca$	$(p)$ The equations represent planes meeting only at a single point.
$(B)$ $a+b+c=0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(q)$ The equations represent the line $x=y=z$.
$(C)$ $a+b+c \neq 0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(r)$ The equations represent identical planes.
$(D)$ $a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ca$	$(s)$ The equations represent the whole of the three-dimensional space.

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