Q.Find the equation of the plane passing through the line of intersection of planes $\bar{r}.(2\hat{i}+\hat{j}-\hat{k})=3$ and $\bar{r}.(5\hat{i}-3\hat{j}+4\hat{k})=-9$ and parallel to the line $\bar{r}=(\hat{i}+3\hat{j}+5\hat{k})+\lambda(2\hat{i}+4\hat{j}+5\hat{k})$.

Write the family of planes through the line of intersection, then use the parallel-to-line condition (normal $\perp$ line direction) to find the family parameter.

The family of planes through the line of intersection of $\bar r\cdot(2\hat i+\hat j-\hat k)=3$ and $\bar r\cdot(5\hat i-3\hat j+4\hat k)=-9$ is

$\left[\bar r\cdot(2\hat i+\hat j-\hat k)-3\right] + k\left[\bar r\cdot(5\hat i-3\hat j+4\hat k)+9\right] = 0$

$\bar r\cdot\left[(2+5k)\hat i+(1-3k)\hat j+(-1+4k)\hat k\right] = 3-9k \quad\text{...(1)}$

This plane's normal is $\bar n = (2+5k)\hat i+(1-3k)\hat j+(-1+4k)\hat k$.

The plane is parallel to the line $\bar r=(\hat i+3\hat j+5\hat k)+\lambda(2\hat i+4\hat j+5\hat k)$, so $\bar n$ must be perpendicular to the line's direction vector $\bar b=(2,4,5)$:

$\bar n\cdot\bar b = 0$

$2(2+5k)+4(1-3k)+5(-1+4k) = 0$

$4+10k+4-12k-5+20k = 0$

$3+18k = 0 \ \Rightarrow\ k=-\dfrac16$

Substitute $k=-\dfrac16$ into (1): …