Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector which is perpendicular to $(\vec{a}+\vec{b}+\vec{c})$ and is represented as $\vec{d}=x(\vec{a}\times\vec{b})+y(\vec{b}\times\vec{c})+z(\vec{c}\times\vec{a})$. Then:
Step-by-Step Solution
Key Concept: The perpendicularity condition $\vec{d} \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$ combined with the scalar triple product properties yields the constraint $x+y+z = 0$.
Since $\vec{d} \perp (\vec{a}+\vec{b}+\vec{c})$, we have $\vec{d} \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$. Substituting $\vec{d}=x(\vec{a}\times\vec{b})+y(\vec{b}\times\vec{c})+z(\vec{c}\times\vec{a})$ and expanding using the scalar triple product property $(\vec{p}\times\vec{q})\cdot\vec{r} = [\vec{p},\vec{q},\vec{r}]$, we get: $x[\vec{a}\times\vec{b},\vec{a}+\vec{b}+\vec{c}] + y[\vec{b}\times\vec{c},\vec{a}+\vec{b}+\vec{c}] + z[\vec{c}\times\vec{a},\vec{a}+\vec{b}+\vec{c}] = 0$. This simplifies to $x[\vec{a},\vec{b},\vec{c}] + y[\vec{a},\vec{b},\vec{c}] + z[\vec{a},\vec{b},\vec{c}] = 0$ (where $[\cdot,\cdot,\cdot]$ denotes scalar triple product). Since vectors are non-coplanar, $[\vec{a},\vec{b},\vec{c}] \neq 0$, giving $x+y+z = 0$. Using the identity for $x+y+z=0$: we have $x^3+y^3+z^3 = 3xyz$ (option 1 is correct) and $xy+yz+zx \leq 0$ follows from $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) = 0$ with $x^2+y^2+z^2 \geq 0$, thus $xy+yz+zx \leq 0$ (option 2 is correct).
Correct Answer: 1,2