Given $\vec a+\vec b+\vec c=\vec 0$ where $\vec a,\vec b,\vec c$ are unit vectors.

Find $\lambda$

We square the given relation: $|\vec a+\vec b+\vec c|^2=0$ $|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$

Since $\vec a,\vec b,\vec c$ are unit vectors, $|\vec a|^2=|\vec b|^2=|\vec c|^2=1$ so $1+1+1+2\lambda=0$ $3+2\lambda=0$ $\lambda=-\frac{3}{2}$

Find $\vec d$

We have $\vec d=\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a$

From $\vec c=-(\vec a+\vec b)$ substitute into the expression:

First, $\vec b\times\vec c=\vec b\times[-(\vec a+\vec b)]$ $=-\vec b\times\vec a-\vec b\times\vec b$ $=-\vec b\times\vec a=\vec a\times\vec b$ since $\vec b\times\vec b=\vec 0$ and $\vec b\times\vec a=-\vec a\times\vec b$.

Next, $\vec c\times\vec a=[-(\vec a+\vec b)]\times\vec a$ $=-\vec a\times\vec a-\vec b\times\vec a$ $=-\vec b\times\vec a=\vec a\times\vec b$

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

Ordered pair $\left(\lambda,\vec d\right)=\left(-\frac{3}{2},3\vec a\times\vec b\right)$

Match with options This is Option C.

Comparison with stored answer Stored correct answer: C

Our derived answer also gives C, so they agree.