Value of $\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{i} \cdot(\hat{j} \times \hat{j})+\hat{k} \cdot(\hat{j} \times \hat{i})+\hat{i} \cdot(\hat{k} \times \hat{j})$ is . . . . . . .

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 the points whose position vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $-i + j + 2k,$ and $4i + 5j + \lambda k$ lie on a plane,then $\lambda = $

If $a, b, c$ are three non-coplanar vectors,then $\frac{a \cdot (b \times c)}{c \times a \cdot b} + \frac{b \cdot (a \times c)}{c \cdot (a \times b)} = $

If $2 \hat{i}-\hat{j}+3 \hat{k}$, $-12 \hat{i}-\hat{j}-3 \hat{k}$, $-\hat{i}+2 \hat{j}-4 \hat{k}$, and $\lambda \hat{i}+2 \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then $\lambda=$

If the vectors $\overrightarrow{a}=\hat{i}+a \hat{j}+a^{2} \hat{k}$,$\overrightarrow{b}=\hat{i}+b \hat{j}+b^{2} \hat{k}$ and $\overrightarrow{c}=\hat{i}+c \hat{j}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$,then the value of $abc$ is