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

For how many distinct real values of $\lambda$ are the vectors $-\lambda^2 \hat{i} + \hat{j} + \hat{k}$,$\hat{i} - \lambda^2 \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} - \lambda^2 \hat{k}$ coplanar?

$A$ unit vector $\vec{e} = a \hat{i} + b \hat{j} + c \hat{k}$ is coplanar with the vectors $\hat{i} - 3 \hat{j} + 5 \hat{k}$ and $3 \hat{i} + \hat{j} - 5 \hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$, then $2 a^2 + 3 b^2 + 4 c^2 =$

If $a, b, c$ are any three coplanar unit vectors,then

If the vectors $i+3j-2k$,$2i-j+4k$ and $3i+2j+xk$ are coplanar,then the value of $x$ 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 $\vec{c} = (0, 1, 0)$: