$f(x)=\max(7-x,x+3)=\frac{|7-x+x+3|+|7-x-x-3|}{2}=\frac{|10|+|4-2x|}{2}=5+|2-x|\\ \therefore f(x)=5+|x-2|$
Now $f(x)$ will be minimum if $|x-2|$ is minimum.
For the option A, it is $-6\le x< -2 \Rightarrow -8\le x -2 < -4 \space \therefore |x-2| \in (4,8] $
For the option B, it is $-2\le x< 2 \Rightarrow -4\le x -2 < 0 \space \therefore |x-2| \in (0,4] $
For the option C, it is $2\le x< 6 \Rightarrow 0\le x -2 < 4 \space \therefore |x-2| \in [0,4) $
For the option D, it is $6\le x< 10 \Rightarrow 4\le x -2 < 8 \space \therefore |x-2| \in [4,8) $
Clearly, for the option C, the value of $|x-2|$ is minimum as $|x-2| \in [0,4)$. So the answer is C.