Let $\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$ and $\vec{b} \cdot \vec{c}=0$. Consider the following two statements:

(A) $|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$ for all $\lambda \in \mathbb{R}$.

(B) $\vec{a}$ and $\vec{c}$ are always parallel.

Then,

$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$ <br/><br/>$\Rightarrow$ $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$ <br/><br/>$$ \begin{aligned} & \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\ & =|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{a}) \\\\ & \Rightarrow \vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=0 \Rightarrow \vec{c} \cdot \vec{a}=0 \\\\ & |\vec{a}+\lambda \vec{c}|^{2}=|\vec{a}|^{2}+\lambda^{2}|\vec{c}|^{2}+0 \geq|\vec{a}|^{2} \end{aligned} $$ <br/><br/>So $\mathrm{A}$ is correct. <br/><br/>$B$ is incorrect.