Let $$\overrightarrow{a}= 2\widehat{i}-5\widehat{j}+5\widehat{k}$$ and $$\overrightarrow{b}= \widehat{i}-\widehat{j}+3\widehat{k}$$. If $$\overrightarrow{C}$$ is a vector such that $$2(\overrightarrow{a}\times\overrightarrow{c})+3(\overrightarrow{b}\times\overrightarrow{c})= \overrightarrow{0}$$ and $$(\overrightarrow{a}-\overrightarrow{b})\cdot\overrightarrow{c}=-97,$$ then $$\mid \overrightarrow{c}\times\widehat{k} \mid^{2}$$ is equal to

We need to find $$|\vec{c} \times \hat{k}|^2$$ given the conditions on vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$.

The vectors are given by $$\vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k}$$ and $$\vec{b} = \hat{i} - \hat{j} + 3\hat{k}$$, and it is known that $$2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$$ as well as $$(\vec{a} - \vec{b}) \cdot \vec{c} = -97$$.

The cross product relation can be written as $$(2\vec{a} + 3\vec{b}) \times \vec{c} = \vec{0}$$, which means $$\vec{c}$$ is parallel to $$2\vec{a} + 3\vec{b}$$.

Calculating $$2\vec{a} + 3\vec{b}$$ gives $$2\vec{a} + 3\vec{b} = 2(2\hat{i} - 5\hat{j} + 5\hat{k}) + 3(\hat{i} - \hat{j} + 3\hat{k}) = (4+3)\hat{i} + (-10-3)\hat{j} + (10+9)\hat{k} = 7\hat{i} - 13\hat{j} + 19\hat{k}$$, so we may write $$\vec{c} = t(7\hat{i} - 13\hat{j} + 19\hat{k})$$ for some scalar $$t$$.

Using $$(\vec{a} - \vec{b}) \cdot \vec{c} = -97$$ and noting that $$\vec{a} - \vec{b} = (2-1)\hat{i} + (-5+1)\hat{j} + (5-3)\hat{k} = \hat{i} - 4\hat{j} + 2\hat{k}$$, we get $$(\vec{a} - \vec{b}) \cdot \vec{c} = t(7 + 52 + 38) = 97t = -97$$, hence $$t = -1$$ and $$\vec{c} = -7\hat{i} + 13\hat{j} - 19\hat{k}$$.

To compute $$|\vec{c} \times \hat{k}|^2$$, note that $$\vec{c} \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{vmatrix} = \hat{i}(13 \cdot 1 - (-19) \cdot 0) - \hat{j}((-7)(1) - (-19)(0)) + \hat{k}(0 - 0) = 13\hat{i} + 7\hat{j}$$, so $$|\vec{c} \times \hat{k}|^2 = 13^2 + 7^2 = 169 + 49 = 218$$.

Therefore, $$|\vec{c} \times \hat{k}|^2 = $$ Option 3: 218.