The distance of the point whose position vector is \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})\) from the plane \(\mathbf{r} \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=4\) is

Equation of plane \(\mathbf{r} \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=4\)
\[
\begin{aligned}
& \Rightarrow \quad \mathbf{r} \cdot \mathbf{n}=d \\
& \text { Point } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& \mathbf{n}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, d=4 \\
& \text { Distance }=\frac{|\mathbf{a} \cdot \mathbf{n}-d|}{|\mathbf{n}|}=\left|\frac{(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})-4}{\sqrt{1^2+(-2)^2+4^2}}\right| \\
& =\left|\frac{2-2-4-4}{\sqrt{1+4+16}}\right|=\left|\frac{-8}{\sqrt{21}}\right|=\frac{8}{\sqrt{21}} \\
&
\end{aligned}
\]