Let $$|\vec{a}|=2,|\vec{b}|=3$$ and the angle between the vectors $$\vec{a}$$ and $$\vec{b}$$ be $$\frac{\pi}{4}$$. Then $$|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}$$ is equal to :

Let for a triangle $$\mathrm{ABC}$$,

$$\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}$$

$$\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$$

$$\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$$

If $$\delta > 0$$ and the area of the triangle $$\mathrm{ABC}$$ is $$5 \sqrt{6}$$, then $$\overrightarrow{C B} \cdot \overrightarrow{C A}$$ is equal to

Let $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$. If a vector $$\vec{d}$$ satisfies $$\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{d} \cdot \vec{a}=24$$, then $$|\vec{d}|^{2}$$ is equal to :