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(D) To determine the nature of $	riangle ABC$ or the collinearity of points $A, B, C$,we calculate the side lengths using the distance formula betwee

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(D) To determine the nature of $	riangle ABC$ or the collinearity of points $A, B, C$,we calculate the side lengths using the distance formula between position vectors: $d = |\vec{P_2} - \vec{P_1}|$.$A$. Given $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$.$|AB| = |(3-2)\hat{i} + (4-3)\hat{j} + (2-4)\hat{k}| = |\hat{i} + \hat{j} - 2\hat{k}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$.$|BC| = |(4-3)\hat{i} + (2-4)\hat{j} + (3-2)\hat{k}| = |\hat{i} - 2\hat{j} + \hat{k}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.$|CA| = |(2-4)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k}| = |-2\hat{i} + \hat{j} + \hat{k}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}$.Since $|AB| = |BC| = |CA| = \sqrt{6}$,$	riangle ABC$ is an equilateral triangle. Thus,$A-I$.$B$. Given $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$.$|AB| = |2\hat{i} + 2\hat{j} + 4\hat{k}| = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6}$.$|BC| = |-6\hat{i} - 6\hat{j} - 12\hat{k}| = \sqrt{36+36+144} = \sqrt{216} = 6\sqrt{6}$.$|CA| = |-4\hat{i} - 4\hat{j} - 8\hat{k}| = \sqrt{16+16+64} = \sqrt{96} = 4\sqrt{6}$.Since $|AB| + |CA| = 2\sqrt{6} + 4\sqrt{6} = 6\sqrt{6} = |BC|$,the points are collinear. Thus,$B-IV$.$C$. Given $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$.$|AB| = |-\hat{i} - 2\hat{j} - 6\hat{k}| = \sqrt{1+4+36} = \sqrt{41}$.$|BC| = |-4\hat{i} - \hat{j} + \hat{k}| = \sqrt{16+1+1} = \sqrt{18}$.$|CA| = |-5\hat{i} - 3\hat{j} - 5\hat{k}| = \sqrt{25+9+25} = \sqrt{59}$.$|AB|^2 + |BC|^2 = 41 + 18 = 59 = |CA|^2$. Thus,$	riangle ABC$ is a right-angled triangle. Thus,$C-III$.$D$. Given $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$.$|AB| = |0\hat{i} + \hat{j} + 2\hat{k}| = \sqrt{0+1+4} = \sqrt{5}$.$|BC| = |\hat{i} - 3\hat{j} - 2\hat{k}| = \sqrt{1+9+4} = \sqrt{14}$.$|CA| = |\hat{i} - 2\hat{j} + 0\hat{k}| = \sqrt{1+4+0} = \sqrt{5}$.Since $|AB| = |CA| = \sqrt{5}$,$	riangle ABC$ is an isosceles triangle. Thus,$D-II$.

List-$I$	List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$	$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$	$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$	$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$	$IV$. $A, B, C$ are collinear

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