Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}+\lambda \hat{k}$ be the co-terminal edges

of a parallelopiped whose volume is 5 units. Then the value of $\lambda$ is

 Volume condition: $|\vec a \cdot (\vec b \times \vec c)| = 5$ 

 Compute cross product using determinant: \[ \vec b \times \vec c = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & 1 & \lambda \end{vmatrix} \] \[ = (2\lambda + 1)\mathbf{i} - (\lambda + 3)\mathbf{j} - 5\mathbf{k} \] So $\vec b \times \vec c = \langle 2\lambda + 1,\; -(\lambda + 3),\; -5\rangle$ 

 Scalar triple product: \[ \vec a \cdot (\vec b \times \vec c) = 2(2\lambda + 1) + (-3)(-(\lambda + 3)) + 4(-5) \] Simplifying: \[ = 4\lambda + 2 + 3(\lambda + 3) - 20 = 7\lambda - 9 \] Volume equation: \[ |7\lambda - 9| = 5 \] Solve: \[ 7\lambda - 9 = 5 \Rightarrow \lambda = 2 \] \[ 7\lambda - 9 = -5 \Rightarrow \lambda = \frac{4}{7} \] Final answer: \[ \boxed{\lambda = 2 \text{ or } \frac{4}{7}} \]