The planes $\overrightarrow{r} \cdot \left(2 \hat{i} + \lambda \hat{j} - 3 \hat{k}\right) = 10$ $\;$ and $\;$ $\overrightarrow{r} \cdot \left(\lambda \hat{i} + 3 \hat{j} + \hat{k}\right) = 5$ $\;$ are perpendicular. Find $\lambda$.
The equations of the given planes are
$\overrightarrow{r} \cdot \left(2 \hat{i} + \lambda \hat{j} - 3 \hat{k}\right) = 10$ $\;\;\; \cdots \; (1a)$
$\overrightarrow{r} \cdot \left(\lambda \hat{i} + 3 \hat{j} + \hat{k}\right) = 5$ $\;\;\; \cdots \; (2a)$
The normals to the given planes are
$\overrightarrow{n_1} = 2 \hat{i} + \lambda \hat{j} - 3 \hat{k}$ $\;\;\; \cdots \; (1b)$
$\overrightarrow{n_2} = \lambda \hat{i} + 3 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$
Let the angle between the planes be $\theta$.
Now, $\cos \theta = \dfrac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right| \left|\overrightarrow{n_2}\right|}$
$\because$ $\;$ The two planes are perpendicular, $\theta = \dfrac{\pi}{2}$ $\;$ and $\;$ $\cos \theta = 0$
$\implies$ $\overrightarrow{n_1} \cdot \overrightarrow{n_2} = 0$
i.e. $\;$ $\left(2 \hat{i} + \lambda \hat{j} - 3 \hat{k}\right) \cdot \left(\lambda \hat{i} + 3 \hat{j} + \hat{k}\right) = 0$
i.e. $\;$ $2 \lambda + 3 \lambda - 3 = 0$ $\implies$ $\lambda = \dfrac{3}{5}$