Correct option: D.Both $(A)$ and $(B)$ are true
d
Without loss of generality
Let $\left|a_1\right| \leq\left|a_2\right| \leq\left|a_3\right|$
$|\vec{a}|^2=\left|a_1\right|^2+\left|a_2\right|^2+\left|a_3\right|^2 \geq\left(a_3\right)^2$
$|\vec{a}| \geq\left|a_3\right|=\max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\}$
$A$ is true
$|\vec{a}|^2=\left| a _1\right|^2+\left| a _2\right|^2+\left| a _3\right|^2 \leq\left| a _3\right|^2+\left| a _3\right|^2+\left| a _3\right|^2$
$|\overrightarrow{ a }|^2 \leq 3\left| a _3\right|^2$
$|\overrightarrow{ a }| \leq \sqrt{3}\left| a _3\right|=\sqrt{3} \max \left\{\left| a _1\right|,\left| a _2\right|,\left| a _3\right|\right\}$
$\leq 3 \max \left\{\left| a _1\right|,\left| a _2\right|,\left| a _3\right|\right\}$
$(2)$ is true