Let vec{a}=hat{i}+hat{j}+hat{k},vec{b}=-hat{i | vedclass

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(A) Given $\vec{b} 	imes \vec{a} = \vec{c} 	imes \vec{a}$,we have $(\vec{b} - \vec{c}) 	imes \vec{a} = 0$.This implies $\vec{b} - \vec{c} = \lambda

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$38$

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(A) Given $\vec{b} 	imes \vec{a} = \vec{c} 	imes \vec{a}$,we have $(\vec{b} - \vec{c}) 	imes \vec{a} = 0$.This implies $\vec{b} - \vec{c} = \lambda \vec{a}$ for some scalar $\lambda$.Substituting the vectors: $(-\hat{i} - 8\hat{j} + 2\hat{k}) - (4\hat{i} + c_2\hat{j} + c_3\hat{k}) = \lambda(\hat{i} + \hat{j} + \hat{k})$.Comparing components:$-1 - 4 = \lambda \implies \lambda = -5$.$-8 - c_2 = \lambda \implies -8 - c_2 = -5 \implies c_2 = -3$.$2 - c_3 = \lambda \implies 2 - c_3 = -5 \implies c_3 = 7$.Thus,$\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$.Let $\vec{d} = 3\hat{i} + 4\hat{j} + \hat{k}$. Then $\cos 	heta = \frac{\vec{c} \cdot \vec{d}}{|\vec{c}| |\vec{d}|} = \frac{(4)(3) + (-3)(4) + (7)(1)}{\sqrt{16+9+49} \sqrt{9+16+1}} = \frac{12 - 12 + 7}{\sqrt{74} \sqrt{26}} = \frac{7}{\sqrt{1924}} = \frac{7}{2\sqrt{481}}$.$\cos^2 	heta = \frac{49}{4 	imes 481} = \frac{49}{1924}$.$	an^2 	heta = \sec^2 	heta - 1 = \frac{1}{\cos^2 	heta} - 1 = \frac{1924}{49} - 1 = \frac{1875}{49} \approx 38.265$.The greatest integer less than or equal to $	an^2 	heta$ is $\lfloor 38.265 floor = 38$.

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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