The equation of the plane containing the line $\vec{r} = \hat{i}+\hat{j} + λ(2\hat{i} + \hat{j} + 4\hat{k})$ is

$\vec{r}.(-\hat{i} -2 \hat{j} + \hat{k})=3$ $\vec{r}.(\hat{i} +2 \hat{j} - \hat{k})=0$ $\vec{r}.(\hat{i} +2 \hat{j} - \hat{k})=3$ none of these

$\vec{r}.(\hat{i} +2 \hat{j} - \hat{k})=3$

Clearly, the position vector of any point on the given line satisfies the equation of the plane $\vec{r}.(\hat{i} +2 \hat{j} - \hat{k})=3$ Hence, the given line lies in the plane given in option (c).