Answer
The set $\{(2,-4,2),(0,2,4),(-10,-4,2)\}$ is orthogonal, not orthonormal and it is a basis for $R^3$.
Work Step by Step
given $\{(2,-4,2),(0,2,4),(-10,-4,2)\}$
let $v_1=(2,-4,2),v_2=(0,2,4),v_3=(-10,-4,2)$
(a)
since
$v_1v_2=0-8+8=0$
$v_1v_3=-20+16+4=0$
$v_2v_3=0-8+8=0$
then, the set $\{(2,-4,2),(0,2,4),(-10,-4,2)\}$ is orthogonal.
(b)
scince
$\left\|{v}_{1}\right\|=\sqrt{v_{1} \cdot v_{1}}=\sqrt{4+16+4}=\sqrt{24}\neq1$
then, the set $\{(4,-1,1),(-1,0,4),(-4,-17,-1)\}$ is not orthonormal.
(c)
by the corollary to Theorem 5.10, $\{(2,-4,2),(0,2,4),(-10,-4,2)\}$ is a basis for $R^3$.
