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(B) Given that $B = 3\hat{i}-2\hat{j}+\hat{k}$ and $C = 5\hat{i}+\hat{j}-3\hat{k}$.Vector $\vec{BC} = C - B = (5-3)\hat{i} + (1-(-2))\hat{j} + (-3-1)\

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$29$

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(B) Given that $B = 3\hat{i}-2\hat{j}+\hat{k}$ and $C = 5\hat{i}+\hat{j}-3\hat{k}$.Vector $\vec{BC} = C - B = (5-3)\hat{i} + (1-(-2))\hat{j} + (-3-1)\hat{k} = 2\hat{i} + 3\hat{j} - 4\hat{k}$.Since $\max \{AB, BC, AC\} = BC$,$BC$ is the hypotenuse of the right-angled triangle $	riangle ABC$,which implies $\angle A = 90^{\circ}$.Therefore,$\vec{AB} \cdot \vec{AC} = 0$.We need to evaluate $\vec{AB} \cdot \vec{AC} + \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$.Since $\vec{AB} \cdot \vec{AC} = 0$,the expression becomes $\vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$.Using the projection formula in $	riangle ABC$,we have $\vec{BA} \cdot \vec{BC} = |\vec{BA}| |\vec{BC}| \cos B = |\vec{BA}|^2$ and $\vec{CA} \cdot \vec{CB} = |\vec{CA}| |\vec{CB}| \cos C = |\vec{CA}|^2$.Thus,the expression is $|\vec{BA}|^2 + |\vec{CA}|^2$.By Pythagoras theorem,$|\vec{BA}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$.$|\vec{BC}|^2 = (2)^2 + (3)^2 + (-4)^2 = 4 + 9 + 16 = 29$.

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

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