Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=2,|\vec{b},|=3|\vec{c}|=5 $ and each one of three vectors is perpendicular to the sum of the other two, then value of $|\vec{a}+\vec{b}+\vec{c}|$ is :

$\sqrt{38}$

The correct answer is Option (2) → $\sqrt{38}$

$\vec a.(\vec b+\vec c)=\vec b.(\vec a+\vec c)=\vec c.(\vec b+\vec a)=0$

so adding them

$\vec a.\vec b+\vec b.\vec c+\vec a.\vec c=0$

so $\sqrt{(\vec a+\vec b+\vec c).(\vec a+\vec b+\vec c)}=|\vec a+\vec b+\vec c|$

$\sqrt{\vec a^2+\vec b^2+\vec c^2+2(\vec a.\vec b+\vec b.\vec c+\vec a.\vec c)}$

$=\sqrt{2^2+3^2+5^2+0}=\sqrt{4+25+9}$

$=\sqrt{38}$