The position vectors of the points $A, B, C$ and $D$ are $3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}-3 \hat{j}+2 \hat{k}, \hat{i}-\hat{j}+2 \hat{k}$ and $4 \hat{i}-\hat{j}-\lambda \hat{k}$ respectively. If the points $A, B, C$ and $D$ lie on a plane, the value of $\lambda$ is

If $\vec{a}, \vec{b}, \vec{c}$ are non-zero and non-coplanar vectors such that $(\vec{a} + \lambda \vec{b}) \cdot [(\vec{b} + 3\vec{c}) \times (\vec{c} - 4\vec{a})] = 0$,then $\lambda$ is equal to

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 the four points with position vectors $-2\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, $\hat{j}-\hat{k}$, and $\lambda\hat{j}+\hat{k}$ are coplanar, then $\lambda$ is equal to

If the points having the position vectors $-\hat{i}+4 \hat{j}-4 \hat{k}$,$3 \hat{i}+2 \hat{j}-5 \hat{k}$,$-3 \hat{i}+8 \hat{j}-5 \hat{k}$ and $-3 \hat{i}+2 \hat{j}+\lambda \hat{k}$ are coplanar,then $\lambda=$

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and the points $P_1 = \lambda \vec{a}+3 \vec{b}-\vec{c}$, $P_2 = \vec{a}-\lambda \vec{b}+3 \vec{c}$, $P_3 = 3 \vec{a}+4 \vec{b}-\lambda \vec{c}$, and $P_4 = \vec{a}-6 \vec{b}+6 \vec{c}$ are coplanar, then one of the values of $\lambda$ is