$$ \text{Let }\overrightarrow{c} \text{ be the projection vector of }\overrightarrow{b}=\lambda\widehat{i}+4\widehat{k}, \lambda > 0 , \text{ on the vector } \overrightarrow{a}=\widehat{i}+2\widehat{j}+2\widehat{k}. \text{ If } \mid \overrightarrow{a}+ \overrightarrow{c}\mid= 7, \text{ then the area of the parallelogram formed by the vectors }\overrightarrow{b} \text{ and }\overrightarrow{c} \text{ is } $$______

$$\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$$, $$\vec{b} = \lambda\hat{i} + 4\hat{k}$$ ($$\lambda > 0$$). $$\vec{c}$$ is the projection vector of $$\vec{b}$$ on $$\vec{a}$$. $$|\vec{a} + \vec{c}| = 7$$.

The projection of $$\vec{b}$$ onto $$\vec{a}$$ is:

$$\vec{c} = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}$$

$$\vec{b} \cdot \vec{a} = \lambda(1) + 0(2) + 4(2) = \lambda + 8$$

$$|\vec{a}|^2 = 1 + 4 + 4 = 9$$

$$\vec{c} = \frac{\lambda + 8}{9} \vec{a} = \frac{\lambda + 8}{9}(\hat{i} + 2\hat{j} + 2\hat{k})$$

$$\vec{a} + \vec{c} = \left(1 + \frac{\lambda+8}{9}\right)\vec{a} = \frac{9 + \lambda + 8}{9}\vec{a} = \frac{\lambda + 17}{9}\vec{a}$$

$$|\vec{a} + \vec{c}| = \frac{|\lambda + 17|}{9} |\vec{a}| = \frac{\lambda + 17}{9} \times 3 = \frac{\lambda + 17}{3}$$

(Since $$\lambda > 0$$, $$\lambda + 17 > 0$$)

$$\frac{\lambda + 17}{3} = 7 \implies \lambda = 4$$

$$\vec{b} = 4\hat{i} + 4\hat{k}$$

$$\vec{c} = \frac{4+8}{9}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{12}{9}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{4}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$$

$$\vec{c} = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j} + \frac{8}{3}\hat{k}$$

Area = $$|\vec{b} \times \vec{c}|$$

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & 4 \\ 4/3 & 8/3 & 8/3 \end{vmatrix}$$

$$= \hat{i}(0 \cdot \frac{8}{3} - 4 \cdot \frac{8}{3}) - \hat{j}(4 \cdot \frac{8}{3} - 4 \cdot \frac{4}{3}) + \hat{k}(4 \cdot \frac{8}{3} - 0)$$

$$= \hat{i}(-\frac{32}{3}) - \hat{j}(\frac{16}{3}) + \hat{k}(\frac{32}{3})$$

$$|\vec{b} \times \vec{c}| = \frac{1}{3}\sqrt{32^2 + 16^2 + 32^2} = \frac{1}{3}\sqrt{1024 + 256 + 1024} = \frac{1}{3}\sqrt{2304} = \frac{48}{3} = 16$$

The area is 16.