Express a general vector $(x, y, z)$ in $R^3$ as a linear combination of $(1, 2, 1),(1, 0, -1),(1, -2, 1)$ $a + b + c = x$
$2a + 0 - 2c = y$
$a – b + c = z$
Then, $(a + b + c, 2a - 2c, a - b + c) = (x, y, z)$, but this looks horrendous unless they mean I should simplify this somehow or something entirely different. Am I not understanding the question correctly?

hint: $a+c=\dfrac{x+z}{2}, a-c = \dfrac{y}{2}\Rightarrow 2a = \dfrac{x+y+z}{2}\Rightarrow a = \dfrac{x+y+z}{4}$, and similarly $b = \dfrac{x-z}{2}$. Can you solve for $c$ ?
