In triangle ABC,if left|begin{array}{lll}a & | vedclass

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(B) The given determinant is $\Delta = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$. Since $a, b, c$ are sides of a triangle,$a+b+c 
eq 0$. Thus,$a^2+b^2+c^2-

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$\frac{3}{4}$

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(B) The given determinant is $\Delta = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$. Since $a, b, c$ are sides of a triangle,$a+b+c 
eq 0$. Thus,$a^2+b^2+c^2-ab-bc-ca = 0$,which implies $a=b=c$. Since $a=b=c$,the triangle is equilateral,so $A=B=C=60^\circ$. Then $\cos A \cos B + \cos B \cos C + \cos C \cos A = \cos 60^\circ \cos 60^\circ + \cos 60^\circ \cos 60^\circ + \cos 60^\circ \cos 60^\circ$. $= \frac{1}{2} 	imes \frac{1}{2} + \frac{1}{2} 	imes \frac{1}{2} + \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.

In $\triangle ABC$,if $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=0$,then $\cos A \cos B+\cos B \cos C+\cos C \cos A=$

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