Explanation

(a)
$\overrightarrow{r}.{\overrightarrow{n}}_{1} + \lambda\overrightarrow{r}.{\overrightarrow{n}}_{2} = q_{1} + \lambda q_{2}$ (i)
Where $\lambda$ is a parameter
So, ${\overrightarrow{n}}_{1} + \lambda{\overrightarrow{n}}_{2}$ is normal to plane (i). Now, any plane parallel to the line of intersection of the planes
$\overrightarrow{r}.{\overrightarrow{n}}_{3} = q_{3}\$and$\ \overrightarrow{r}.{\overrightarrow{n}}_{4} = q_{4}$ is of form $\overrightarrow{r}.({\overrightarrow{n}}_{3} \times {\overrightarrow{n}}_{4}) = d$. Hence we must have
$\left\lbrack {\overrightarrow{n}}_{1} + \lambda{\overrightarrow{n}}_{2} \right\rbrack.\lbrack{\overrightarrow{n}}_{3} \times {\overrightarrow{n}}_{4}\rbrack = 0$
$\Rightarrow \left\lbrack {\overrightarrow{n}}_{1}\overrightarrow{n_{3}}{\overrightarrow{n}}_{4} \right\rbrack + \lambda\left\lbrack {\overrightarrow{n}}_{2}{\overrightarrow{n}}_{3}{\overrightarrow{n}}_{4} \right\rbrack = 0$
$\Rightarrow \ \lambda = \frac{- \lbrack{\overrightarrow{n}}_{1}\overrightarrow{n_{3}}{\overrightarrow{n}}_{4}\rbrack}{\lbrack{\overrightarrow{n}}_{2}{\overrightarrow{n}}_{3}{\overrightarrow{n}}_{4}\rbrack}$
$\Rightarrow$ On putting this value in Eq. (i), we have the equation of the required plane as
$\overrightarrow{r}.{\overrightarrow{n}}_{1} - q_{1} = \frac{\left\lbrack {\overrightarrow{n}}_{1}\overrightarrow{n_{3}}{\overrightarrow{n}}_{4} \right\rbrack}{\left\lbrack {\overrightarrow{n}}_{2}{\overrightarrow{n}}_{3}{\overrightarrow{n}}_{4} \right\rbrack}(r.\ {\overrightarrow{n}}_{2} - q_{2})$
$\Rightarrow \ \left\lbrack {\overrightarrow{n}}_{2}{\overrightarrow{n}}_{3}{\overrightarrow{n}}_{4} \right\rbrack\left( \overrightarrow{r}.{\overrightarrow{n}}_{1} - q_{1} \right) = \left\lbrack {\overrightarrow{n}}_{1}\overrightarrow{n_{3}}{\overrightarrow{n}}_{4} \right\rbrack(\overrightarrow{r}.{\overrightarrow{n}}_{2} - q_{2})$