The projection of \(\vec{b}\) onto \(\vec{a}\) is \(\vec{c}=\frac{\vec{b}\cdot\vec{a}}{|\vec{a}|^2}\vec{a}\). We have \(\vec{b}\cdot\vec{a}=\lambda(1)+0(2)+4(2)=\lambda+8\) and \(|\vec{a}|^2=1+4+4=9\). So \(\vec{c}=\frac{\lambda+8}{9}(\hat{i}+2\hat{j}+2\hat{k})\). Then \(\vec{a}+\vec{c}=\left(1+\frac{\lambda+8}{9}\right)\hat{i}+2\left(1+\frac{\lambda+8}{9}\right)\hat{j}+2\left(1+\frac{\lambda+8}{9}\right)\hat{k}=\frac{17+\lambda}{9}(\hat{i}+2\hat{j}+2\hat{k})\). Its magnitude: \(|\vec{a}+\vec{c}|=\frac{17+\lambda}{9}\cdot 3=\frac{17+\lambda}{3}=7\Rightarrow\lambda=4\). Now \(\vec{b}=4\hat{i}+4\hat{k}\) and \(\vec{c}=\frac{12}{9}(\hat{i}+2\hat{j}+2\hat{k})=\frac{4}{3}\hat{i}+\frac{8}{3}\hat{j}+\frac{8}{3}\hat{k}\). Area of parallelogram \(=|\vec{b}\times\vec{c}|\). Computing the cross product and its magnitude gives \(|\vec{b}\times\vec{c}|=16\).