Gọi A(a,0,0) là điểm cần tìm
\(d\left(A,\left(\alpha\right)\right)=\frac{\left|a+0-0-1\right|}{\sqrt{1^2+1^2+1^2}}=\frac{\left|a-1\right|}{\sqrt{3}}\)
\(d\left(B,\left(\beta\right)\right)=\frac{\left|2a+0+2.0-2\right|}{\sqrt{2^2+1^2+2^2}}=\frac{\left|2a-2\right|}{3}\)
Ta có \(d\left(A,\left(\beta\right)\right)=d\left(B,\left(\beta\right)\right)\)\(\Leftrightarrow\frac{\left|a-1\right|}{\sqrt{3}}=\frac{\left|2a-2\right|}{3}\\ \Leftrightarrow\left|a-1\right|=0\\ \Leftrightarrow a=1\)