Given vectors $\vec a=\hat i+2\hat j+2\hat k=(1,2,2),\qquad \vec b=\lambda \hat i+4\hat k=(\lambda,0,4),\ \lambda>0$

Projection vector of $\vec b$ on $\vec a$ The projection of $\vec b$ on $\vec a$ is $\vec c=\operatorname{proj}_{\vec a}\vec b=\frac{\vec b\cdot \vec a}{|\vec a|^2}\vec a$

First compute: $\vec b\cdot \vec a=\lambda(1)+0(2)+4(2)=\lambda+8$ $|\vec a|^2=1^2+2^2+2^2=9$

Hence, $\vec c=\frac{\lambda+8}{9}(1,2,2)$

Use the condition $|\vec a+\vec c|=7$ Since $\vec c$ is parallel to $\vec a$, write $\vec c=t\vec a,\qquad t=\frac{\lambda+8}{9}$ Then $\vec a+\vec c=(1+t)\vec a$ So, $|\vec a+\vec c|=|1+t|\,|\vec a|=7$

Now, $|\vec a|=3$ Therefore, $|1+t|\cdot 3=7\quad\Rightarrow\quad |1+t|=\frac73$

Since $\lambda>0$, we have $t=\frac{\lambda+8}{9}>0$ so $1+t>0$. Hence, $1+t=\frac73\Rightarrow t=\frac43$

Thus, $\frac{\lambda+8}{9}=\frac43$ $\lambda+8=12\Rightarrow \lambda=4$

Now find $\vec b$ and $\vec c$ $\vec b=(4,0,4)$ $\vec c=\frac43\vec a=\frac43(1,2,2)=\left(\frac43,\frac83,\frac83\right)$

Area of parallelogram formed by $\vec b$ and $\vec c$ Area = $|\vec b\times \vec c|$.

Compute:

Final answer The area of the parallelogram is $\boxed{16}$