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Let $P = n(n+1) (2n+1)$

Now, we can check all of the option.

- It is even

Let’s take various values of $n$ and check.

- $n = 1: P = 1 (2)(3) = 6$
- $n = 2: P = 2 (3)(5) = 30$
- $n = 3: P = 3 (4) (7) = 84$

$\qquad \downarrow$

- $P = n[\text{Odd}] \; (n+1)[\text{Even}] \; (2n+1)[\text{always Odd}] = \text{Even} \quad [\text{Odd} \times \text{Even = Even}]$

So, option $A$ is correct.

- Divisible by $3.$

Let’s take various of $n$ and check.

- $n = 1: P = 1 (2)(3) = 6 \Rightarrow \text{Yes}$
- $n = 2: P = 2 (3)(5) = 30 \Rightarrow \text{Yes}$
- $n = 3: P = 3 (4) (7) = 84 \Rightarrow \text{Yes}$

$\qquad \downarrow$

- $P = n(n+1) (2n+1) \Rightarrow \text{Yes}$

So, option $B$ is correct.

Divisible by the sum of the square of square of first $n$ natural numbers.

The sum of the square of first $n$ natural numbers $ = 1^{2} + 2^{2} + 3^{2} + n^{2} = \frac{n(n+1)(2n+1)}{6}$

$ = \frac{P}{6}$

So, $\frac{P}{\frac{P}{6}} = 6.$

So, option $C$ is correct.

- Never divisible by $237.$

Let take $n = 236$

$P = n(n+1) (2n+1)$

$P = 236 (237) [2(236)+1]$

It is divisible by $237$

We can also take $n = 118$

$P = 118 (119)(237)$

So, option $D$ is not correct.

Correct Answer$: \text{D}$