The equation of the plane through \((-1,1,2)\) whose normal makes equal acute angles with co-ordinate axes is

Note that \((-1,1,2)\) is satisfied by only option (B) Alternate Method:
Let \(A \equiv(-1,1,2)\)
\(\therefore \bar{a} =-\hat{i}+\hat{j}+2 \hat{k} \)
\( \bar{n}=\hat{i}+\hat{j}+\hat{k}\)
\(\therefore\) equation of plane is \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}\)
\(
\begin{aligned}
& \Rightarrow \overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& \Rightarrow \overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2 \\
& \Rightarrow(x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2 \\
& \Rightarrow x+y+\mathrm{z}-2=0
\end{aligned}
\)