Solution:-

let us assume the three consecutive natural number be $P – 1$, $P$ and $P + 1$

As per the condition given in the question, \[{{\left( P-1 \right)}^{2}}~+\text{ }\left( P \right)\text{ }\left( P\text{ }+\text{ }1 \right)\text{ }=\text{ }154\]

Then,

\[{{\left( P-1 \right)}^{2}}~+\text{ }{{P}^{2}}~+\text{ }P\text{ }=\text{ }154\]

We know that,

\[{{\left( a-b \right)}^{2}}~=\text{ }{{a}^{2}}-2ab\text{ }+\text{ }{{b}^{2}}\]

\[{{P}^{2}}-2P\text{ }+\text{ }1\text{ }+\text{ }{{P}^{2}}~+\text{ }P\text{ }=\text{ }154\]

\[2{{P}^{2}}-P\text{ }+\text{ }1\text{ }=\text{ }154\]

By transposing we get,

\[2{{P}^{2}}-P\text{ }+\text{ }1-154\text{ }=\text{ }0\]

\[2{{P}^{2}}-P-153\text{ }=\text{ }0\]

\[2{{P}^{2}}-18P-17P-153\text{ }=\text{ }0\]

Take out common in each terms,

$2P(P – 9) – 17(P – 9) = 0$

$(P – 9) (2P – 17) = 0$

Equate both to zero,

$P – 9 = 0$

$2P – 17 = 0$

$P = 9$

$2P = 17$

$P = 9$

$P = 17/2$

So, $P = 9$ … [because $9$ is a natural number]

Then,

$P = 9$

$P – 1 = 9 – 1 = 8$

$P + 1 = 9 + 1= 10$

Therefore, the three consecutive natural number are $8$, $9$ and $10$.