Explanation

(a)
Given that $x,\ y,\mathcal{z}\ \in R$ and distinct and
$u =$ $x^{2} + 4y^{2} + 9\mathcal{z}^{2} - 6y\mathcal{z -}3\mathcal{z}x - 2xy$
$= \frac{1}{2}(2x^{2} + 8y^{2} + 18\mathcal{z}^{2} - 12y\mathcal{z -}6\mathcal{z}x - 4xy)$
$= \frac{1}{2}\{\left( x^{2} - 4xy + 4y^{2} \right) + \left( x^{2} - 6x\mathcal{z} + 9\mathcal{z}^{2} \right) + \left( 4y^{2} - 12y\mathcal{z} + 9\mathcal{z}^{2} \right)\}$
$= \frac{1}{2}\left\{ (x - 2y)^{2} + \left( x - 3\mathcal{z} \right)^{2} + \left( 2y - 3\mathcal{z} \right)^{2} \right\} > 0$
So, $u$ is always non-negative.