$$\vec{a}=4 \hat{i}+13 \hat{j}-18 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$$ and $$\vec{c}=2 \hat{i}+3 \hat{j}-4 \hat{k}$$ are three vectors such that $$\vec{a}=x \vec{b}+y \vec{c}$$, then $$x+y=$$

If $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=\hat{j}-\hat{k}, \vec{a} \times \bar{b}=\bar{c}$$ and $$\vec{a} \cdot \vec{b}=1$$, then $$\vec{b}$$

If the vectors $$\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}$$ and $$\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}$$ are collinear, then $$p$$ and $$q$$ are

If $$|\vec{a}|=4,|\vec{b}|=5$$, then the values of $$k$$ for which $$\vec{a}+k \vec{b}$$ is perpendicular to $$\vec{a}-k \vec{b}$$ are