If $\bar{a} = \bar{i} - \bar{j}$,$\bar{b} = \bar{j} - \bar{k}$,$\bar{c} = \bar{k} - \bar{i}$ and $\bar{d}$ is a unit vector such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$,then the vector $\bar{d} = ....$

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$:

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{a} \times \vec{b} = \vec{c}$,$\vec{b} \times \vec{c} = \vec{a}$ and $|\vec{a}| = 2$. Then which one of the following is not true?

If $\overline{a}=\hat{i}-\hat{k}$,$\overline{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\overline{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $\overline{a} \cdot(\overline{b} \times \overline{c})$ depends on

The points $A(4, 5, 1)$,$B(0, -1, -1)$,$C(3, 9, 4)$,and $D(-4, 4, 4)$ are:

If $\overline{a}$ and $\overline{c}$ are unit vectors inclined at $\frac{\pi}{3}$ with each other and $(\overline{a} \times (\overline{b} \times \overline{c})) \cdot (\overline{a} \times \overline{c}) = 5$,then the value of $5[\overline{a} \overline{b} \overline{c}] = $