Correct Answer: D \((2)\) The solution is found by using the formula for the volume of a parallelepiped, which is the absolute value of the scalar triple product of its co-terminal edge vectors. Step 1: The Formula for the Volume of a Parallelepiped The volume \(V\) of a parallelepiped with co-terminal edges given by vectors \(\vec a,\vec b,\vec c\) is \[ V=\left|\vec a\cdot(\vec b\times \vec c)\right| =\left|\begin{vmatrix} a_x & a_y & a_z\\ b_x & b_y & b_z\\ c_x & c_y & c_z \end{vmatrix}\right|. \] Step 2: Set Up and Calculate the Determinant Given vectors: \[ \vec a=2\hat i-3\hat j+4\hat k \implies (2,-3,4), \] \[ \vec b=\hat i+2\hat j-\hat k \implies (1,2,-1), \] \[ \vec c=3\hat i+\hat j+\lambda \hat k \implies (3,1,\lambda). \] Compute: \[ \begin{vmatrix} 2 & -3 & 4\\ 1 & 2 & -1\\ 3 & 1 & \lambda \end{vmatrix} = 2\begin{vmatrix} 2 & -1\\ 1 & \lambda \end{vmatrix} -(-3)\begin{vmatrix} 1 & -1\\ 3 & \lambda \end{vmatrix} +4\begin{vmatrix} 1 & 2\\ 3 & 1 \end{vmatrix}. \] \[ =2(2\lambda-(-1))+3(\lambda-(-3))+4(1-6) \] \[ =2(2\lambda+1)+3(\lambda+3)+4(-5) \] \[ =4\lambda+2+3\lambda+9-20 \] \[ =7\lambda-9. \] Step 3: Solve for \(\lambda\) using the Given Volume Given \(V=5\), \[ |7\lambda-9|=5. \] So, \[ 7\lambda-9=5 \Rightarrow 7\lambda=14 \Rightarrow \lambda=2, \] or \[ 7\lambda-9=-5 \Rightarrow 7\lambda=4 \Rightarrow \lambda=\frac{4}{7}. \] From the options, \(\lambda=2\) is present, hence it is the correct answer.