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

• A
  -$1$
• B
  $1/2$
• C
  $-3/2$
• D
  $2$