Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$. If $\vec{p}=\vec{a} \times \vec{b}$,then the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$ and $|\vec{p} \times \vec{c}|=3$. Thus,$\vec{a} \cdot \vec{c}$ is equal to:

  • A
    $\frac{1}{8}$
  • B
    $1$
  • C
    $2$
  • D
    $4$

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