$\text{ A line passes through the point }A(\hat{i}+2\hat{j}+3\hat{k})\text{ and is parallel to the vector }\overrightarrow{V}(\hat{i}+\hat{j}+\hat{k})\text{ . }$The shortest distance from the origin, to the line is……

 

$\text{ Equation of line: }\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=t\text{ . }$

$\text{ Let }P=(t+1,t+2,t+3)\text{ be the foot of the perpendicular from origin }$ $\text{ Then }(t+1)\left(1\right)+(t+2)\left(1\right)+(t+3)\left(1\right)=0$

$t=-2$

$\therefore P=\left(-1,0,1\right)andOP=\sqrt{2}$