Let $$\overrightarrow{c} \text{ and } \overrightarrow{d}$$ be vectors such that $$\mid\overrightarrow{c}+\overrightarrow{d}\mid=\sqrt{29}$$ and $$\overrightarrow{c}\times( 2\widehat{i}+3\widehat{j}+4\widehat{k})=(2\widehat{i}+3\widehat{j}+4\widehat{k})\times\overrightarrow{d}$$. If $$\lambda_{1}, \lambda_{2}( \lambda_{1}> \lambda_{2})$$ are the possible values of $$(\overrightarrow{c}+\overrightarrow{d})\cdot(-7\widehat{i}+2\widehat{j}+3\overrightarrow{k})$$, then the equation $$K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+\left(3K+\frac{\lambda_{2}}{2} \right)y^{2}-8x+12y+\lambda_{2}=0$$ represents a circle, for K equal to :

 From the equation $$\vec{c} \times (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k}) \times \vec{d}$$, 

let $$\vec{e} = 2\hat{i}+3\hat{j}+4\hat{k}$$ so that it becomes $$\vec{c} \times \vec{e} = \vec{e} \times \vec{d}$$. Since $$\vec{e} \times \vec{d} = -\vec{d} \times \vec{e}$$, we have $$\vec{c} \times \vec{e} + \vec{d} \times \vec{e} = \vec{0}$$, implying $$(\vec{c} + \vec{d}) \times \vec{e} = \vec{0}$$. 

$$\vec{c} + \vec{d}$$ is parallel to $$\vec{e}$$ and can be written as $$\vec{c} + \vec{d} = \lambda\vec{e}$$ for some scalar $$\lambda$$.

Using the condition $$|\vec{c} + \vec{d}| = \sqrt{29}$$, we get $$|\lambda \vec{e}| = |\lambda|\sqrt{4 + 9 + 16} = |\lambda|\sqrt{29} = \sqrt{29}$$, which gives $$|\lambda| = 1\implies \lambda = \pm1$$.

find $$(\vec{c} + \vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$$, substitute $$\vec{c} + \vec{d} = \lambda\vec{e}$$: $$\lambda\vec{e} \cdot (-7\hat{i}+2\hat{j}+3\hat{k}) = \lambda[2(-7) + 3(2) + 4(3)] = 4\lambda$$.

For $$\lambda = 1$$, the value is 4, and for $$\lambda = -1$$ it is -4. 

Since $$\lambda_1 > \lambda_2$$, we take $$\lambda_1 = 4$$ and $$\lambda_2 = -4$$.

 $$K^2x^2 + (K^2 - 5K + \lambda_1)xy + (3K + \frac{\lambda_2}{2})y^2 - 8x + 12y + \lambda_2 = 0$$ and substitute $$\lambda_1 = 4$$ and $$\lambda_2 = -4$$ to obtain $$K^2x^2 + (K^2 - 5K + 4)xy + (3K - 2)y^2 - 8x + 12y - 4 = 0$$.

For a general second-degree equation $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ to represent a circle, the coefficient of $$x^2$$ must equal that of $$y^2$$ (so $$a=b$$) and the coefficient of $$xy$$ must be zero (so $$h=0$$).

Equating the coefficients gives two conditions: $$K^2 = 3K - 2$$ and $$K^2 - 5K + 4 = 0$$. The first yields $$K^2 - 3K + 2 = 0\implies (K-1)(K-2) = 0$$, so $$K = 1$$ or $$2$$. The second gives $$(K-1)(K-4) = 0$$, so $$K = 1$$ or $$4$$.

The only value of $$K$$ satisfying both conditions is $$K = 1$$.