Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}$,$x \in R$. If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$,then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to

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?

Let $\overrightarrow{PR}=3 \hat{i}+\hat{j}-2 \hat{k}$ and $\overrightarrow{SQ}=\hat{i}-3 \hat{j}-4 \hat{k}$ be the diagonals of a parallelogram $PQRS$,and let $\overrightarrow{PT}=\hat{i}+2 \hat{j}+3 \hat{k}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{PT}, \overrightarrow{PQ}$ and $\overrightarrow{PS}$ is:

If the four points $2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{a} - 2\vec{b} + 3\vec{c}$,$3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{a} - \lambda\vec{b} - 6\vec{c}$ are coplanar,find the value of $\lambda$.

Consider the four points $A(1, -2, -1)$, $B(4, 0, -3)$, $C(1, 2, -1)$, and $D(2, -4, -5)$ in space. If $\vec{b} = \vec{AB}$, $\vec{c} = \vec{AC}$, and $\vec{d} = \vec{AD}$, then find the value of $\frac{[\vec{b} \times \vec{c}, \vec{c} \times \vec{d}, \vec{d} \times \vec{b}]}{[\vec{b}+\vec{c}, \vec{c}+\vec{d}, \vec{d}+\vec{b}]}$.

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 $\lambda=$