Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three unit vectors such that $$\vec{a} + \vec{b} + \vec{c} = 0$$. If $$\lambda = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$$ and $$\vec{d} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$$, then the order pair, $$\left(\lambda, \vec{d}\right)$$, is equal to

We are given three unit vectors $$\vec a ,\; \vec b,\; \vec c$$ satisfying $$\vec a+\vec b+\vec c = 0.$$

First we compute $$\lambda = \vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a.$$ We start by taking the dot product of the equation $$\vec a+\vec b+\vec c = 0$$ with itself. The dot-product formula $$\vec x\cdot\vec x = |\vec x|^{2}$$ gives

$$\left(\vec a+\vec b+\vec c\right)\cdot\left(\vec a+\vec b+\vec c\right)=0.$$

Expanding the left side term by term, we have

$$\vec a\cdot\vec a + \vec b\cdot\vec b + \vec c\cdot\vec c + 2\left(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a\right)=0.$$

Because each vector is a unit vector, $$\vec a\cdot\vec a=\vec b\cdot\vec b=\vec c\cdot\vec c=1.$$ Substituting these values gives

$$1+1+1 + 2\lambda = 0 \quad\Longrightarrow\quad 3 + 2\lambda = 0.$$

Solving for $$\lambda$$ yields

$$\lambda = -\dfrac{3}{2}.$$

Next we find $$\vec d = \vec a\times\vec b + \vec b\times\vec c + \vec c\times\vec a.$$ From $$\vec a+\vec b+\vec c=0$$ we get $$\vec c = -(\vec a+\vec b).$$ We substitute this expression wherever $$\vec c$$ occurs in $$\vec d$$:

$$\vec d = \vec a\times\vec b + \vec b\times\!\bigl(-(\vec a+\vec b)\bigr) + \bigl(-(\vec a+\vec b)\bigr)\times\vec a.$$

Simplifying each cross product one by one:

$$\vec b\times\!\bigl(-(\vec a+\vec b)\bigr) = -\vec b\times\vec a - \vec b\times\vec b = -\vec b\times\vec a - \vec 0 = -\vec b\times\vec a,$$ since $$\vec b\times\vec b=\vec 0.$$

Similarly,

$$( -(\vec a+\vec b) )\times\vec a = -\vec a\times\vec a - \vec b\times\vec a = -\vec 0 - \vec b\times\vec a = -\vec b\times\vec a.$$

Therefore

$$\vec d = \vec a\times\vec b -\vec b\times\vec a -\vec b\times\vec a.$$

The cross product is anti-commutative, so $$\vec b\times\vec a = -\vec a\times\vec b.$$ Using this fact, each of the last two terms becomes

$$-\vec b\times\vec a = -(-\vec a\times\vec b)= \vec a\times\vec b.$$

Thus we obtain

$$\vec d = \vec a\times\vec b + \vec a\times\vec b + \vec a\times\vec b = 3\,\vec a\times\vec b.$$

Combining both results, the ordered pair $$\left(\lambda,\;\vec d\right)$$ equals

$$\left(-\dfrac{3}{2},\; 3\,\vec a\times\vec b\right).$$

Hence, the correct answer is Option D.