Answer
$\textbf{a}=⟨\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3}⟩+⟨\frac{2x-y-z}{3},\frac{-x+2y-z}{3},\frac{-x-y+2z}{3}⟩$
Work Step by Step
We are given the vectors $\textbf{a}=⟨x,y,z⟩$ and $\textbf{b}=⟨1,1,1⟩$. We have,
$\textbf{a}_{\parallel \textbf{b}}=(\frac{\textbf{a}\cdot\textbf{b}}{\textbf{b}\cdot\textbf{b}})\textbf{b}=\frac{x\times1+y\times1+z\times1}{1^{2}+1^{2}+1^{2}}⟨1,1,1⟩=⟨\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3}⟩$
$\textbf{a}_{\perp \textbf{b}}=\textbf{a}-\textbf{a}_{\parallel \textbf{b}}=$
$⟨x,y,z⟩-⟨\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3}⟩$
$=⟨\frac{2x-y-z}{3},\frac{-x+2y-z}{3},\frac{-x-y+2z}{3}⟩$
Therefore, the decomposition of $\textbf{a}$ with respect to $\textbf{b}$ is
$\textbf{a}=\textbf{a}_{\parallel \textbf{b}}+\textbf{a}_{\parallel \textbf{b}}$$=⟨\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3}⟩+⟨\frac{2x-y-z}{3},\frac{-x+2y-z}{3},\frac{-x-y+2z}{3}⟩$
