In triangle ABC,the points P, Q, R divide BC, | vedclass

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(D) Let the position vectors of $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Let $\vec{a} = \vec{0}$. Then $\vec{b} = \vec{b}$ and $\vec{c}

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$3:2$

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(D) Let the position vectors of $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Let $\vec{a} = \vec{0}$. Then $\vec{b} = \vec{b}$ and $\vec{c} = \vec{c}$.Points $P, Q, R$ divide $BC, CA, AB$ in ratios $3:4, 2:5, 9:5$ respectively:$\vec{P} = \frac{3\vec{c} + 4\vec{b}}{7}$,$\vec{Q} = \frac{5\vec{c} + 2\vec{a}}{7} = \frac{5\vec{c}}{7}$,$\vec{R} = \frac{9\vec{b} + 5\vec{a}}{14} = \frac{9\vec{b}}{14}$.Now,$\vec{AP} + \vec{BQ} + \vec{CR} = (\vec{P} - \vec{a}) + (\vec{Q} - \vec{b}) + (\vec{R} - \vec{c})$$= \frac{3\vec{c} + 4\vec{b}}{7} + (\frac{5\vec{c}}{7} - \vec{b}) + (\frac{9\vec{b}}{14} - \vec{c})$$= \frac{6\vec{c} + 8\vec{b} + 10\vec{c} - 14\vec{b} + 9\vec{b} - 14\vec{c}}{14} = \frac{3\vec{b} + 2\vec{c}}{14}$.Point $D$ divides $BC$ in ratio $2:3$,so $\vec{D} = \frac{2\vec{c} + 3\vec{b}}{5}$.Since $\vec{A} = \vec{0}$,$\vec{AD} = \vec{D} = \frac{3\vec{b} + 2\vec{c}}{5}$.Thus,$\frac{\vec{AP} + \vec{BQ} + \vec{CR}}{\vec{AD}} = \frac{(3\vec{b} + 2\vec{c})/14}{(3\vec{b} + 2\vec{c})/5} = \frac{5}{14}$.So,$k = \frac{5}{14}$.Finally,$(14k + 1) : (14k - 1) = (14 	imes \frac{5}{14} + 1) : (14 	imes \frac{5}{14} - 1) = (5 + 1) : (5 - 1) = 6 : 4 = 3 : 2$.

In $\triangle ABC$,the points $P, Q, R$ divide $BC, CA, AB$ in the ratio $3:4, 2:5, 9:5$,respectively,and the point $D$ divides $BC$ in the ratio $2:3$. If $\vec{AP} + \vec{BQ} + \vec{CR} = k \vec{AD}$,then $(14k + 1) : (14k - 1) = $

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