If three vectors $2\hat{i}-\hat{j}-\hat{k}$,$\hat{i}+2\hat{j}-3\hat{k}$ and $3\hat{i}+\lambda\hat{j}+5\hat{k}$ are coplanar,then the value of $\lambda$ is

$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

The volume of the tetrahedron whose coterminus edges are represented by $\bar{a}=-12 \hat{i}+p \hat{k}$,$\bar{b}=3 \hat{j}-\hat{k}$,and $\bar{c}=2 \hat{i}+\hat{j}-15 \hat{k}$ is $570$ cubic units. Then,$p=$

If the vectors $\bar{a}=\hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\bar{b}=2 \hat{\imath}-5 \hat{\jmath}+p \hat{k}$ and $\bar{c}=5 \hat{\imath}-9 \hat{\jmath}+4 \hat{k}$ are coplanar,then the value of $p$ 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 $\vec{c} = (0, 1, 0)$:

Let $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$, $\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$. If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$, then $\lambda$ is equal to