If $a, b, c$ are non-negative distinct numbers and $a \hat{\imath}+a \hat{\jmath}+c \hat{k}$,$\hat{\imath}+\hat{k}$ and $c \hat{\imath}+c \hat{\jmath}+b \hat{k}$ are coplanar vectors,then

If $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{k}$ and $\vec{c}$ is a unit vector, then the maximum value of the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is

Statement $1$: The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ lie in the same plane if and only if $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
Statement $2$: The vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v} = 0$,where $\vec{u} \times \vec{v}$ is a vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$.

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 $\vec{c} = (0, 1, 0)$:

If $\hat{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\hat{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(2 \hat{a}-\hat{b}) \cdot[(\hat{a} \times \hat{b}) \times(\hat{a}+2 \hat{b})]$ is

The vector $(\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k}$ is equal to