Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}$, then which of the following statements are correct?

(A) $|\vec{a}| = \sqrt{14}$

(B) $|\vec{b}| = 29$

(C) $\vec{a} \cdot \vec{b} = 8$

(D) Angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{-8}{\sqrt{406}}\right)$

Choose the correct answer from the options given below:

Given vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}$.

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Checking statement (A): $|\vec{a}| = \sqrt{14}$

For vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, the magnitude is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$

$|\vec{a}| = \sqrt{(1)^2 + (2)^2 + (3)^2}$

$= \sqrt{1 + 4 + 9}$

$= \sqrt{14}$

Statement (A) is correct.

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Checking statement (B): $|\vec{b}| = 29$

$|\vec{b}| = \sqrt{(-2)^2 + (3)^2 + (-4)^2}$

$= \sqrt{4 + 9 + 16}$

$= \sqrt{29}$

Since $\sqrt{29} \neq 29$, statement (B) is incorrect.

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Checking statement (C): $\vec{a} \cdot \vec{b} = 8$

The dot product formula is $\vec{a} \cdot \vec{b} = (x_1)(x_2) + (y_1)(y_2) + (z_1)(z_2)$

$\vec{a} \cdot \vec{b} = (1)(-2) + (2)(3) + (3)(-4)$

$= -2 + 6 - 12$

$= -8$

Since the dot product is $-8$, not $8$, statement (C) is incorrect.

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Checking statement (D): Angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{-8}{\sqrt{406}}\right)$

The angle $\theta$ between two vectors is given by $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Using $\vec{a} \cdot \vec{b} = -8$, $|\vec{a}| = \sqrt{14}$, and $|\vec{b}| = \sqrt{29}$:

$\cos\theta = \frac{-8}{\sqrt{14} \times \sqrt{29}}$

$= \frac{-8}{\sqrt{406}}$

Therefore $\theta = \cos^{-1}\left(\frac{-8}{\sqrt{406}}\right)$

Statement (D) is correct.

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The correct statements are (A) and (D) only.