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(B) $I$. True. Consider a triangle $T_1$ with sides $0.9, 0.9, 0.9 	ext{ cm}$ (equilateral),area $\approx 0.35 	ext{ cm}^2$. Consider a triangle $T_

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Exactly two statements are $INCORRECT$.

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(B) $I$. True. Consider a triangle $T_1$ with sides $0.9, 0.9, 0.9 	ext{ cm}$ (equilateral),area $\approx 0.35 	ext{ cm}^2$. Consider a triangle $T_2$ with base $20 	ext{ m}$ and height $10^{-10} 	ext{ cm}$. Area of $T_2 = \frac{1}{2} 	imes 2000 	ext{ cm} 	imes 10^{-10} 	ext{ cm} = 10^{-7} 	ext{ cm}^2$. Since $0.35 > 10^{-7}$,the statement is true.$II$. Let $a = \frac{1}{x-y}, b = \frac{1}{y-z}, c = \frac{1}{z-x}$. Note that $a+b+c = \frac{(y-z)(z-x) + (x-y)(z-x) + (x-y)(y-z)}{(x-y)(y-z)(z-x)} = 0$. We know $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) = 0$. Thus $a^2+b^2+c^2 = -2(ab+bc+ca)$. The given equation claims $a^2+b^2+c^2 = (a+b+c)^2 = 0$,which is false unless $a=b=c=0$,which is impossible. So,$II$ is $INCORRECT$.$III$. Let $a = \log_3 x, b = \log_4 x, c = \log_5 x$. Equation is $abc = ab+bc+ca$. This is $1/c + 1/a + 1/b = 1$. $\log_x 5 + \log_x 3 + \log_x 4 = 1 \implies \log_x 60 = 1 \implies x = 60$. Also $x=1$ is a solution if we consider the equation as $0=0$ (though $\log 1 = 0$). However,for $x=1$,the terms are defined. The statement says "exactly one",but $x=1$ and $x=60$ are solutions. So,$III$ is $INCORRECT$.$IV$. Factors of $12$ are $(1,12), (2,6), (3,4), (4,3), (6,2), (12,1)$. There are $6$ such orders. This is true.Thus,statements $II$ and $III$ are $INCORRECT$. The answer is $B$.

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