Let the 3 consecutive positive numbers = $$(x-1),(x),(x+1)$$ According to ques, => $$(x-1)^2+(x)^2(x+1)^2=365$$ => $$(x^2-2x+1)+(x^2)+(x^2+2x+1)=365$$ => $$3x^2+2=365$$ => $$3x^2=365-2=363$$ => $$x^2=frac{363}{3}=121$$ => $$x=sqrt{121}=11$$ $$ herefore$$ Sum of numbers = $$(x-1)+(x)+(x+1)=3x$$ = $$3 imes11=33$$ => Ans - (B)