1. Given data

We have $\vec a=2\hat i+\hat j+\hat k,$ and nonzero vectors $\vec b,\vec c$ such that $|\vec a+\vec b+\vec c|=|\vec a+\vec b-\vec c|$ and $\vec b\cdot \vec c=0.$

We must test statements:

• (A) $|\vec a+\lambda \vec c|\ge |\vec a|$ for all $\lambda\in\mathbb R$
• (B) $\vec a$ and $\vec c$ are always parallel

2. Use the condition involving magnitudes

Let $\vec x=\vec a+\vec b.$ Then the given condition becomes $|\vec x+\vec c|=|\vec x-\vec c|.$ Squaring both sides, $|\vec x+\vec c|^2=|\vec x-\vec c|^2.$ So, $(\vec x+\vec c)\cdot(\vec x+\vec c)=(\vec x-\vec c)\cdot(\vec x-\vec c).$ Expanding, $|\vec x|^2+|\vec c|^2+2\vec x\cdot \vec c=|\vec x|^2+|\vec c|^2-2\vec x\cdot \vec c.$ Hence, $4\vec x\cdot \vec c=0 \quad\Rightarrow\quad (\vec a+\vec b)\cdot \vec c=0.$ Thus, $\vec a\cdot \vec c+\vec b\cdot \vec c=0.$ Given $\vec b\cdot \vec c=0$, we get $\boxed{\vec a\cdot \vec c=0.}$

So $\vec a$ is perpendicular to $\vec c$.

3. Check statement (A)

Consider $|\vec a+\lambda \vec c|^2=(\vec a+\lambda \vec c)\cdot(\vec a+\lambda \vec c).$ Expanding, $|\vec a+\lambda \vec c|^2=|\vec a|^2+\lambda^2|\vec c|^2+2\lambda(\vec a\cdot \vec c).$ But $\vec a\cdot \vec c=0$, so $|\vec a+\lambda \vec c|^2=|\vec a|^2+\lambda^2|\vec c|^2\ge |\vec a|^2.$ Therefore, $|\vec a+\lambda \vec c|\ge |\vec a|\quad \text{for all }\lambda\in\mathbb R.$ So (A) is true.

4. Check statement (B)

We found $\vec a\cdot \vec c=0,$ so $\vec a$ and $\vec c$ are perpendicular, not always parallel.

Since $\vec c$ is nonzero and $\vec a=(2,1,1)$ is also nonzero, a vector perpendicular to $\vec a$ certainly exists; for example, $\vec c=\hat i-2\hat j$ gives $\vec a\cdot \vec c=2(1)+1(-2)+1(0)=0.$ This $\vec c$ is clearly not parallel to $\vec a$.

Hence (B) is false.

5. Conclusion

• (A) is correct
• (B) is incorrect

Therefore the correct option is $\boxed{\text{C: only (A) is correct}}.$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.