For any vector $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$, with $10\left|a_{i}\right|<1, i=1,2,3$, consider the following statements :

(A): $$\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\} \leq|\vec{a}|$$

(B) : $$|\vec{a}| \leq 3 \max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\}$$

We have, <br/><br/>$$ \begin{aligned} & 10\left|a_i\right|<1, i=1,2,3 \\\\ & \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\ & |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\ & \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \text {. } \end{aligned} $$ <br/><br/>Hence, (A) is true. <br/><br/>$$ \begin{array}{rlrl} & |\vec{a}| =\sqrt{a_1^2+a_2^2+a_3^2} \leq \sqrt{a_1^2+a_1^2+a_1^2} \\\\ & =\sqrt{3}\left|a_1\right| \\\\ \therefore & |\vec{a}|=\sqrt{3}\left|a_1\right|<3\left|a_1\right| \\\\ \therefore & |\vec{a}|<3 \max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\} \end{array} $$ <br/><br/>Hence, (B) is also true.