Let the position vectors of the points $A, B, C$ and $D$ be $5\hat{i}+5\hat{j}+2\lambda\hat{k}$,$\hat{i}+2\hat{j}+3\hat{k}$,$-2\hat{i}+\lambda\hat{j}+4\hat{k}$ and $-\hat{i}+5\hat{j}+6\hat{k}$. Let the set $S = \{\lambda \in \mathbb{R} : \text{The points } A, B, C \text{ and } D \text{ are coplanar}\}$. Then $\sum_{\lambda \in S}(\lambda+2)^2$ is equal to

Let the volume of the tetrahedron with vertices $\hat{i}-\hat{j}-2\hat{k}$, $-2\hat{i}+\hat{j}-2\hat{k}$, $-\hat{i}-2\hat{j}+\hat{k}$, and $2\hat{i}+2\hat{j}+a\hat{k}$ be $\frac{20}{3}$. Then the integral value of $a$ is

The maximum value and minimum value of the volume of the parallelepiped having coterminous edges $\hat{i}+x \hat{j}+\hat{k}$,$\hat{j}+x \hat{k}$,and $x \hat{i}+\hat{k}$ are respectively:

Let $\vec{p} = 2\hat{i} + 3\hat{j} + a\hat{k}$,$\vec{q} = b\hat{i} + 5\hat{j} - \hat{k}$,and $\vec{r} = \hat{i} + \hat{j} + 3\hat{k}$. If $\vec{p}, \vec{q}, \vec{r}$ are coplanar and $\vec{p} \cdot \vec{q} = 20$,then the ordered pair $(a, b)$ is:

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to:

If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar, then $x=$