If $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ are coterminous edges of a parallelepiped,then its volume is

If $x$ and $y$ are real numbers such that $\hat{i}+\hat{j}+\hat{k}$, $-2 \hat{i}+3 \hat{j}+2 \hat{k}$, $x \hat{i}-5 \hat{j}+3 \hat{k}$, and $\hat{i}+y \hat{j}-\hat{k}$ are the position vectors of four coplanar points, then the locus of $P(x, y)$ is

The lines $\overline{r}=\overline{a}+\lambda(\overline{b} \times \overline{c})$ and $\overline{r}=\overline{c}+\mu(\overline{a} \times \overline{b})$ will intersect if

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:

Let $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$. If $[\vec{u} \, \vec{v} \, \vec{w}] = 0$ and $\vec{w} = \lambda \vec{x} + \mu \vec{y}$ where $(a + b + c) \neq 0$ and $\lambda, \mu \neq 0$,then the vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are:

If the points $2a+3b-c, a-2b+3c, 3a+\lambda b-2c$ and $a-6b+6c$ are coplanar, then the direction cosines of the vector $\lambda \hat{i}-2\lambda \hat{j}+\hat{k}$ are