The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is -

Answer: Option B

Let the 3 consecutive number is
x, x + 1, x + 2
According to question,
$$\Rightarrow x^2+{\left(x+1\right)}^2+{\left(x+2\right)}^2$$     = 365
$$\Rightarrow x^2+x^2+1+2x+x^2+4+4x$$       = 365
$$\begin{array}{rl}& \Rightarrow 3x^2+6x=360\\ & \Rightarrow x^2+2x-120=0\\ & \Rightarrow \left(x-10\right)\left(x+12\right)=0\\ & \Rightarrow x=10\\ & \text{Sum of numbers :}\\ & =10+11+12\\ & =33\end{array}$$