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(B) Given $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$,so $|\overrightarrow{PQ}| = \sqrt{(-2)^2+(-1)^2+2^2} = \sqrt{4+1+4} = 3$.Given $\overrighta

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$37$

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(B) Given $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$,so $|\overrightarrow{PQ}| = \sqrt{(-2)^2+(-1)^2+2^2} = \sqrt{4+1+4} = 3$.Given $\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}$ and $|\overrightarrow{PR}|=9$,so $a^2+b^2+(-4)^2 = 9^2 \implies a^2+b^2+16=81 \implies a^2+b^2=65$ ...$(1)$.Given $\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}$,so $|\overrightarrow{PS}| = \sqrt{1^2+(-7)^2+2^2} = \sqrt{1+49+4} = \sqrt{54} = 3\sqrt{6}$.Since $S$ is equidistant from $PQ$ and $PR$,$PS$ is the angle bisector of $\angle QPR$. Let $\angle QPS = \angle RPS = 	heta$.Then $\cos 	heta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}| |\overrightarrow{PS}|} = \frac{(-2)(1)+(-1)(-7)+(2)(2)}{3 \cdot 3\sqrt{6}} = \frac{-2+7+4}{9\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}}$.Also,$\cos 	heta = \frac{\overrightarrow{PR} \cdot \overrightarrow{PS}}{|\overrightarrow{PR}| |\overrightarrow{PS}|} = \frac{(a)(1)+(b)(-7)+(-4)(2)}{9 \cdot 3\sqrt{6}} = \frac{a-7b-8}{27\sqrt{6}}$.Equating the two expressions for $\cos 	heta$: $\frac{1}{\sqrt{6}} = \frac{a-7b-8}{27\sqrt{6}} \implies a-7b-8 = 27 \implies a-7b = 35$ ...$(2)$.From $(1)$,$a^2+b^2=65$. Substituting $a=35+7b$ into $(1)$: $(35+7b)^2+b^2=65 \implies 1225+490b+49b^2+b^2=65 \implies 50b^2+490b+1160=0 \implies 5b^2+49b+116=0$.Solving for $b$: $b = \frac{-49 \pm \sqrt{49^2-4(5)(116)}}{10} = \frac{-49 \pm \sqrt{2401-2320}}{10} = \frac{-49 \pm \sqrt{81}}{10} = \frac{-49 \pm 9}{10}$.So $b = -4$ or $b = -5.8$. Since $b \in \mathbb{Z}$,$b=-4$.Then $a = 35+7(-4) = 35-28 = 7$.Thus,$3a-4b = 3(7)-4(-4) = 21+16 = 37$.

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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