The cartesian equation of the plane $\overrightarrow{\mathbf{r}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2$ is ............. .
$$\begin{array}{lr} \text { We have, } & \overrightarrow{\mathbf{r}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 \\ \Rightarrow & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 \\ \Rightarrow & x+y-z=2 \end{array}$$
which is the required form
The unit vector normal to the plane $x+2 y+3 z-6=0$ is $$\frac{1}{\sqrt{14}} \hat{\mathbf{i}}+\frac{2}{\sqrt{14}} \hat{\mathbf{j}}+\frac{3}{\sqrt{14}} \hat{\mathbf{k}}.$$
The intercepts made by the plane $2 x-3 y+5 z+4=0$ on the coordinate axis are $-2, \frac{4}{3}$ and $-\frac{4}{5}$.
The angle between the line $\overrightarrow{\mathbf{r}}=(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}-4 \hat{\mathbf{k}})+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$ and the plane $\overrightarrow{\mathbf{r}}(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}})+5=0$ is $\sin ^{-1}\left(\frac{5}{2 \sqrt{91}}\right)$.
The angle between the planes $\overrightarrow{\mathbf{r}}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})=1$ and $\overrightarrow{\mathbf{r}}(\hat{\mathbf{i}}-\hat{\mathbf{j}})=4$ is $$\cos ^{-1}\left(\frac{-5}{\sqrt{58}}\right)$$