Let $$\overrightarrow{a}=-\widehat{i}+2\widehat{j}+2\widehat{k},\overrightarrow{b}=8\widehat{i}+7\widehat{j}-3\widehat{k} \text { and } \overrightarrow{c}$$ be a vector such that $$\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}$$. If $$\overrightarrow{c}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=4$$, then $$\mid\overrightarrow{a}+\overrightarrow{c}\mid^{2}$$ is equal to :

Let $$\vec{c} = (x, y, z)$$. 

Since $$\vec{c} \cdot (1,1,1) = 4$$,

 $$x + y + z = 4$$.

 $$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ x & y & z \end{vmatrix} = (2z - 2y,\; 2x + z,\; -y - 2x)$$

equating to $$(8,7,-3)$$ gives $$2z - 2y = 8,\quad 2x + z = 7,\quad -y - 2x = -3,$$ or equivalently $$z - y = 4,\quad 2x + z = 7,\quad 2x + y = 3.$$

From $$x + y + z = 4$$, the second equation gives $$z = 7 - 2x$$ and the third gives $$y = 3 - 2x$$. 

Substituting into $$x + y + z = 4$$ yields x + (3 - 2x) + (7 - 2x) = 4

$$\Longrightarrow\; x = 2,$$ hence $$y = 3 - 2(2) = -1$$ and $$z = 7 - 2(2) = 3$$. 

Thus $$\vec{c} = (2, -1, 3)$$.

$$\vec{a} + \vec{c} = (-1 + 2,\; 2 - 1,\; 2 + 3) = (1,1,5),$$ so $$|\vec{a}+\vec{c}|^2 = 1^2 + 1^2 + 5^2 = 27.$$