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 :

Let $$a, b, c$$ be three distinct real numbers, none equal to one. If the vectors $$a \hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+b \hat{j}+\hat{\mathrm{k}}$$ and $$\hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$$ are coplanar, then $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$ is equal to :

Let $$\lambda \in \mathbb{Z}, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{c}$$ be a vector such that $$(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$$ and $$\vec{b} \cdot \vec{c}=-20$$. Then $$|\vec{c} \times(\lambda \hat{i}+\hat{j}+\hat{k})|^{2}$$ is equal to :