Direction vectors: \( \vec d_1 = \langle 4,1,1 \rangle,\quad \vec d_2 = \langle 1,1,0 \rangle \) \( \vec d = \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \langle -1,1,3 \rangle \) --- Equation of line \(L\) through \(P(1,1,1)\): \( x=1-t,\; y=1+t,\; z=1+3t \) --- For \(Q\) (on yz-plane ⇒ \(x=0\)): \( t=1 \Rightarrow Q(0,2,4) \) --- Line through \(S(1,0,-1)\) parallel to \(L\): \( x=1-\mu,\; y=\mu,\; z=-1+3\mu \) --- For \(R\) (on yz-plane ⇒ \(x=0\)): \( \mu=1 \Rightarrow R(0,1,2) \) --- Vectors: \( \vec{PQ}=\langle -1,1,3 \rangle,\quad \vec{PS}=\langle 0,-1,-2 \rangle \) --- Area of parallelogram: \( \vec{PQ}\times \vec{PS} = \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 1 & 3 \\ 0 & -1 & -2 \end{vmatrix} = \langle 1,-2,1 \rangle \) --- Area: \( |\vec{PQ}\times \vec{PS}| = \sqrt{1^2+(-2)^2+1^2} = \sqrt{6} \) --- Required square: \( = 6 \)