For a triangle ABC, let $$\overrightarrow{p} = \overrightarrow{BC}, \overrightarrow{q}= \overrightarrow{CA}$$ and $$\overrightarrow{r} = \overrightarrow{BA}$$. If $$|\overrightarrow{p}| = 2\sqrt{3}, |\overrightarrow{q}|=2$$ and $$\cos\theta = \frac{1}{\sqrt{3}}$$ where $$\theta$$ is the angle between $$\overrightarrow{p}$$ and $$\overrightarrow{q}$$, then $$|\overrightarrow{p} \times \left(\overrightarrow{q}-\overrightarrow{3r}\right)|^2 +3|\overrightarrow{r}|^2$$ is equal to :

Given $$\vec{p} = \vec{BC}$$, $$\vec{q} = \vec{CA}$$, $$\vec{r} = \vec{BA}$$, $$|\vec{p}| = 2\sqrt{3}$$, $$|\vec{q}| = 2$$, and $$\cos\theta = 1/\sqrt{3}$$ where $$\theta$$ is the angle between $$\vec{p}$$ and $$\vec{q}$$.

Note that $$\vec{r} = \vec{BA} = \vec{BC} + \vec{CA} = \vec{p} + \vec{q}$$.

$$|\vec{r}|^2 = |\vec{p}|^2 + 2\vec{p}\cdot\vec{q} + |\vec{q}|^2 = 12 + 2\vec{p}\cdot\vec{q} + 4$$

$$\vec{p}\cdot\vec{q} = |\vec{p}||\vec{q}|\cos\theta = 2\sqrt{3} \cdot 2 \cdot \frac{1}{\sqrt{3}} = 4$$

$$|\vec{r}|^2 = 12 + 8 + 4 = 24$$

Compute $$\vec{q} - 3\vec{r} = \vec{q} - 3(\vec{p} + \vec{q}) = -3\vec{p} - 2\vec{q}$$.

$$\vec{p} \times (\vec{q} - 3\vec{r}) = \vec{p} \times (-3\vec{p} - 2\vec{q}) = -3(\vec{p} \times \vec{p}) - 2(\vec{p} \times \vec{q}) = -2(\vec{p} \times \vec{q})$$

$$|\vec{p} \times \vec{q}|^2 = |\vec{p}|^2|\vec{q}|^2\sin^2\theta = 12 \cdot 4 \cdot (1 - 1/3) = 48 \cdot \frac{2}{3} = 32$$

$$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 = 4|\vec{p} \times \vec{q}|^2 = 4 \times 32 = 128$$

Final answer:

$$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2 = 128 + 3 \times 24 = 128 + 72 = 200$$

The answer is Option 4: 200.