Let the vectors $$\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$$ and $$\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$$. Then, the set of all values of $$t$$ is :

Let a vector $$\vec{a}$$ has magnitude 9. Let a vector $$\vec{b}$$ be such that for every $$(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$$, the vector $$(x \vec{a}+y \vec{b})$$ is perpendicular to the vector $$(6 y \vec{a}-18 x \vec{b})$$. Then the value of $$|\vec{a} \times \vec{b}|$$ is equal to :

Let $$\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$$ and $$\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$$ be two vectors, such that $$\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$$. Then the projection of $$\vec{b}-2 \vec{a}$$ on $$\vec{b}+\vec{a}$$ is equal to :

$$ \text { Let } \vec{a}=2 \hat{i}-\hat{j}+5 \hat{k} \text { and } \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k} \text {. If }((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2} \text {, then }|\vec{b} \times 2 \hat{j}| $$ is equal to :