Find a vector of length 1 satisfying given relations

Given vectors $A = (1,-2,3)$ and $B = (3,1,2)$ in $\mathbb{R}^3$ find a vector $C$ of unit length parallel to the following vectors:

$A+B$ ;
$A-B$ ;
$A+2B$ ;
$A-2B$ ;
$2A-B$ .

First, we have
$A+ B = (4,-1,5).$

Since $C$ must be parallel to this we know there is a nonzero scalar $x$ such that

$C = (4x,-x,5x).$

Since $C$ is unit length we have

$\lVert C \rVert = 1 \quad \implies \quad \sqrt{16x^2 +x^2 + 25x^2} = 1 \quad \implies \quad x = \pm \frac{1}{\sqrt{42}}.$

Therefore, $C = \frac{1}{42}(4,-1,5)$ .
First, we have
$A- B = (-2,-3,1).$

Since $C$ must be parallel to this we know there is a nonzero scalar $x$ such that

$C = (-2x,-3x,x).$

Since $C$ is unit length we have

$\lVert C \rVert = 1 \quad \implies \quad \sqrt{4x^2 + 9x^2 + x^2} = 1 \quad \implies \quad x = \pm \frac{1}{\sqrt{14}}.$

Therefore, $C = \frac{1}{\sqrt{14}}(-2,-3,1)$ .
First, we have
$A+2B = (7,0,7).$

Since $C$ must be parallel to this we know there is a nonzero scalar $x$ such that

$C = (x,0,x).$

Since $C$ is unit length we have

$\lVert C \rVert = 1 \quad \implies \quad \sqrt{2x^2} = 1 \quad \implies \quad x = \pm \frac{1}{\sqrt{2}}.$

Therefore, $C = \frac{1}{\sqrt{2}} (1,0,1)$ .
First, we have
$A - 2B = (-5,-4,-1).$

Since $C$ must be parallel to this we know there is a nonzero scalar $x$ such that

$C = (-5x,-4x,-x).$

Since $C$ is unit length we have

$\lVert C \rVert = 1 \quad \implies \quad \sqrt{25x^2 + 16x^2 + x^2} = 1 \quad \implies \quad x = \pm \frac{1}{\sqrt{42}}.$

Therefore, $C = \frac{1}{\sqrt{42}} (-5,-,4,-1)$ .
First, we have
$2A - B = (-1,-5,4).$

Since $C$ must be parallel to this we know there is a nonzero scalar $x$ such that

$C = (-x,-5x,4x).$

Since $C$ is unit length we have

$\lVert C \rVert = 1 \quad \implies \quad \sqrt{x^2 + 25x^2 + 16x^2} = 1 \quad \implies \quad x = \pm \frac{1}{\sqrt{42}}.$

Therefore, $C = \frac{1}{\sqrt{42}} (-1,-5,4)$ .