Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})$$. If $$(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$$, then $$|\vec{c}|^2$$ is equal to:

Let $$\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$ and $$\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$$ be three vectors. Let $$\overrightarrow{\mathrm{r}}$$ be a unit vector along $$\vec{b}+\vec{c}$$. If $$\vec{r} \cdot \vec{a}=3$$, then $$3 \lambda$$ is equal to:

The set of all $$\alpha$$, for which the vectors $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$ and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is

Let $$\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$$. Then the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is: