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(A) Given position vectors: $\vec{OA} = \vec{a}+\vec{b}+\vec{c}$ and $\vec{OB} = \vec{a}-2\vec{b}+3\vec{c}$.Vector $\vec{AB} = \vec{OB} - \vec{OA} = (

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$4|PQ|$

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(A) Given position vectors: $\vec{OA} = \vec{a}+\vec{b}+\vec{c}$ and $\vec{OB} = \vec{a}-2\vec{b}+3\vec{c}$.Vector $\vec{AB} = \vec{OB} - \vec{OA} = (\vec{a}-2\vec{b}+3\vec{c}) - (\vec{a}+\vec{b}+\vec{c}) = -3\vec{b}+2\vec{c}$.Point $P$ divides $AB$ internally in ratio $1:3$:$\vec{OP} = \frac{3\vec{OA} + 1\vec{OB}}{1+3} = \frac{3(\vec{a}+\vec{b}+\vec{c}) + (\vec{a}-2\vec{b}+3\vec{c})}{4} = \frac{4\vec{a}+\vec{b}+6\vec{c}}{4}$.Point $Q$ divides $AB$ externally in ratio $1:3$:$\vec{OQ} = \frac{3\vec{OA} - 1\vec{OB}}{3-1} = \frac{3(\vec{a}+\vec{b}+\vec{c}) - (\vec{a}-2\vec{b}+3\vec{c})}{2} = \frac{2\vec{a}+5\vec{b}}{2}$.Vector $\vec{PQ} = \vec{OQ} - \vec{OP} = \frac{2\vec{a}+5\vec{b}}{2} - \frac{4\vec{a}+\vec{b}+6\vec{c}}{4} = \frac{4\vec{a}+10\vec{b} - 4\vec{a}-\vec{b}-6\vec{c}}{4} = \frac{9\vec{b}-6\vec{c}}{4} = \frac{3}{4}(3\vec{b}-2\vec{c})$.Since $\vec{AB} = -3\vec{b}+2\vec{c}$,we have $\vec{PQ} = -\frac{3}{4}\vec{AB}$.Taking magnitudes,$|PQ| = \frac{3}{4}|AB|$,which implies $4|PQ| = 3|AB|$.

Let the position vectors of two points $A$ and $B$ be $\vec{a}+\vec{b}+\vec{c}$ and $\vec{a}-2\vec{b}+3\vec{c}$,respectively. If the points $P$ and $Q$ divide $AB$ in the ratio $1:3$ internally and externally respectively,then $3|AB|=$

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