If $\vec a,\vec b,\vec c$ are three mutually perpendicular unit vectors, then $|\vec a + \vec b + \vec c|$ is equal to

$\sqrt{3}$

The correct answer is Option (3) → $\sqrt{3}$

Given $\vec a,\vec b,\vec c$ are mutually perpendicular unit vectors

$|\vec a|=|\vec b|=|\vec c|=1$ and $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$

$|\vec a+\vec b+\vec c|^2=(\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)$

$=|\vec a|^2+|\vec b|^2+|\vec c|^2$

$=1+1+1$

$=3$

$|\vec a+\vec b+\vec c|=\sqrt3$

The value of $|\vec a+\vec b+\vec c|$ is $\sqrt3$.