Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 292. Which are those three integers?
A. 6, 7, 8
B. 4, 5, 6
C. 7, 8, 9
D. 5, 6, 7

Hint: We will be using the concepts of integers and number systems to solve the problem. We will be converting the condition given in the question into an equation and solving it to get an answer.

Complete step by step answer:
So we have been three consecutive integers. So let $n-1,n,n+1$ be three consecutive integers.
Now it has been given that the square of their sum exceeds the sum of their square by 292. So, we will now write this in algebraic equation as
${{\left( n-1+n+n+1 \right)}^{2}}={{\left( n-1 \right)}^{2}}+{{n}^{2}}+{{\left( n+1 \right)}^{2}}+292$
${{\left( 3n \right)}^{2}}={{\left( n-1 \right)}^{2}}+{{n}^{2}}+{{\left( n+1 \right)}^{2}}+292$
Now, we know that
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Therefore,
$9{{n}^{2}}={{n}^{2}}+1-2n+{{n}^{2}}+{{n}^{2}}+1+2n+292$
$9{{n}^{2}}=3{{n}^{2}}+294$
$6{{n}^{2}}=294$
${{n}^{2}}=49$
$n=\pm 7$
Now by taking $n=7$ we have the set of consecutive integers as
$7-1,7,7+1$
$\Rightarrow 6,7,8$

So, the correct answer is “Option A”.

Note: To solve this type of question it is important to convert the conditions given in the question into an equation then subsequently solve it or answer. Here we need to avoid negative values as already given positive integers.