If $$\bar{a}=\hat{\boldsymbol{i}}-\hat{\boldsymbol{k}}, \bar{b}=x \hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+(1-x) \hat{\boldsymbol{k}}$$ and $$\bar{c}=y \hat{\boldsymbol{i}}+x \hat{\boldsymbol{j}}+(1+x-y) \hat{\boldsymbol{k}}$$, then $$[\bar{a} \bar{b} \bar{c}]$$ depends on

Let $$\bar{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$$ and $$\bar{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$$ be two vectors. If $$\bar{c}$$ is a vector such that $$\bar{b} \times \bar{c}=\bar{b} \times \bar{a}$$ and $$\bar{c} \cdot \bar{a}=0$$, then $$\bar{c} \cdot \bar{b}$$ is equal to

The magnitude of the projection of the vector $$2 \hat{\mathbf{i}}+ 3\hat{\mathbf{j}}+\hat{\mathbf{k}}$$ on the vector perpendicular to the plane containing the vectors $$\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$$ and $$\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$$ is

If $$\bar{a}=\hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}, \bar{b}=\hat{\boldsymbol{i}}-\hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}$$ and $$\bar{c}=\hat{\boldsymbol{i}}-\hat{\boldsymbol{j}}-\hat{\boldsymbol{k}}$$ are three vectors then vector $$\bar{r}$$ in the plane of $$\bar{a}$$ and $$\bar{b}$$, whose projection on $$\bar{c}$$ is $$\frac{1}{\sqrt{3}}$$, is given by