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

$[a, b, a \times b]$ is equal to

If $a, b, c$ are non-coplanar vectors and $d = \lambda a + \mu b + \nu c$,then $\lambda$ is equal to

$i \cdot(j \times k)+j \cdot(k \times i)+k \cdot(j \times i)$ is equal to

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 $\vec{c} = (0, 1, 0)$:

If $\vec a = 2\sin \theta \hat i - \hat j + 2\hat k$,$\vec b = 2\hat i + 2\sin \theta \hat j - \hat k$ and $\vec c = 4\hat i + \hat j + 4\cos^2 \theta \hat k$ are coplanar,then $\theta$ can be equal to