The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,then the vector equation of the line is

  • A
    $\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+5 \hat{k})$
  • B
    $\bar{r}=(2 \hat{i}-4 \hat{j}-5 \hat{k})+\lambda(-3 \hat{i}+2 \hat{j}-5 \hat{k})$
  • C
    $\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(10 \hat{i}+25 \hat{j}-16 \hat{k})$
  • D
    $\bar{r}=(3 \hat{i}+2 \hat{j}+5 \hat{k})+\lambda(10 \hat{i}+25 \hat{j}-16 \hat{k})$

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