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 $$\cos ^{-1}\left(\frac{-5}{\sqrt{58}}\right)$$

The line $\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ lies in the plane $\overrightarrow{\mathbf{r}}(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+2=0$.

The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is $\overrightarrow{\mathbf{r}}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}+\lambda(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$

The equation of a line, which is parallel to $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and which passes through the point $(5,-2,4)$ is $\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}$.

If the foot of perpendicular drawn from the origin to a plane is $(5,-3,-2)$, then the equation of plane is $\overrightarrow{\mathbf{r}}(5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=38$.