If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}-\bar{b}|^2+|\bar{b}-\bar{c}|^2+|\bar{c}-\bar{a}|^2=15$,then $|\bar{a}-\bar{b}-\bar{c}|^2-4(\bar{b} \cdot \bar{c})=$

  • A
    $6$
  • B
    $15$
  • C
    $12$
  • D
    $10$

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Let $a, b, c$ be three vectors such that the magnitude of $b$ is twice that of $a$ and the magnitude of $c$ is three times that of $a$. If the angle between each pair of vectors is $\frac{\pi}{3}$ and $|a+b+c|=5$, then $|c|+|a|+|b|=$

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