I) $$ab(a + b) + bc(b + c) + ca(c +a) \leq 6abc $$

$$\frac{ab(a+b)}{abc}+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}\le6$$

$$\frac{(a+b)}{c}+\frac{(b+c)}{a}+\frac{(c+a)}{b}\leq6$$

We shall rearrange the terms

$$\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{a}\le6$$

Here, we have six terms: three fractions and their reciprocals.

$$t+\frac{1}{t}\ge2$$ this is always true.

So three such equations add to $$\ge6$$

So, it is only sometimes valid.

II. $$\frac{a^{2} + b^{2} + c^{2}}{abc} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$

$$\frac{a^2+b^2+c^2}{abc}\le\frac{a\cdot b+b\cdot c+c\cdot a}{abc}$$

$$a^2+b^2+c^2\le a\cdot b+b\cdot c+c\cdot a$$

Consider (a,b,c) as (1,2,3)

Substituting them in the above equation,

$$1^2+2^2+3^2\le1\cdot2+2\cdot3+3\cdot1$$

$$14\le11$$ is not true, so this is also not true.

Hence, option D is correct.