Do some computations with vectors in R3

Given vectors

$A = (1,1,1), \qquad B = (0,1,1), \qquad C = (2,1,1)$

in $\mathbb{R}^3$ , let $D = xA + yB + zC$ where $x,y,z \in \mathbb{R}$ are scalars.

Compute the components of $D$ .
Find values of $x,y,z$ such that $D \neq O$ and at least one of $x,y,z$ is nonzero.
Prove that there is no triple $x,y,z$ such that $D = (1,2,3)$ .

We compute,
$D = x(1,1,1) + y(0,1,1) + z(2,1,1) = (x + 2z, x+y+z, x+y+z).$
Proof. Let $x = 2, \ y = -1, \ z = -1$ , then we have
$D = (x+2z,x+y+z,x+y+z) = (2-2, 2 -1-1, 2-1-1) = (0,0,0) = O.$

Hence, these are three values of $x,y,z$ , at least one of which is nonzero, such that $D = O. \qquad \blacksquare$
Proof. We know the components of $D$ are
$D = (2x + z, x+y+z, x+y+z).$

But, if $D = (1,2,3)$ this would implies $x + y + z = 2$ and $x+y+z = 3$ . This is impossible since $2 \neq 3. \qquad \blacksquare$