If the planes $\bar{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\bar{r} \cdot(4 \hat{i}-\hat{j}+\mu \hat{k})=5$ are parallel, then $\lambda+\mu=$

The position vector forms of the planes are $\bar{r} \cdot (2\hat{i} - \lambda\hat{j} + \hat{k}) = 3$ and $\bar{r} \cdot (4\hat{i} - \hat{j} + \mu\hat{k}) = 5$, with normal vectors $\bar{n_1} = 2\hat{i} - \lambda\hat{j} + \hat{k}$ and $\bar{n_2} = 4\hat{i} - \hat{j} + \mu\hat{k}$ respectively.

For planes to be parallel, their normals must be parallel, implying direction proportionality:

$\frac{2}{4} = \frac{-\lambda}{-1} = \frac{1}{\mu}$

Since $\frac{2}{4} = \frac{1}{2}$, equating the remaining ratios yields:

$\frac{-\lambda}{-1} = \frac{1}{2} \Rightarrow \lambda = \frac{1}{2}$

$\frac{1}{\mu} = \frac{1}{2} \Rightarrow \mu = 2$

The sum is $\lambda + \mu = \frac{1}{2} + 2 = \frac{5}{2}$.