If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are the position vectors of the points $A, B, C, D$ respectively such that $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$,then the position vector of the point of intersection of the line segments $AC$ and $BD$ is
- A$\frac{\bar{b}+3 \bar{d}}{4}$
- B$\frac{3 \bar{a}+2 \bar{c}}{5}$
- C$\frac{\bar{a}+\bar{c}}{2}$
- D$\frac{\bar{b}+4 \bar{d}}{5}$
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