Let $$\overrightarrow{AB} = 2\widehat{i}+4\widehat{j}-5\widehat{k}$$ and $$ \overrightarrow{AD} = \widehat{i}+2\widehat{j}+\lambda\widehat{k}, \lambda\text{ }\epsilon \text{ } R$$. Let the projection of the vector $$ \overrightarrow{v}=\widehat{i}+\widehat{j}+\widehat{k}$$ on the disgonal $$\overrightarrow{AC}$$ of the parallelogram ABCD be of length one unit. If $$\alpha> \beta$$, be the roots of the equation $$\lambda^{2}x^{2}-6\lambda x+5=0$$, then $$2\alpha-\beta$$ is equal to

$$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD} = (2\widehat{i} + 4\widehat{j} - 5\widehat{k}) + (\widehat{i} + 2\widehat{j} + \lambda\widehat{k}) = 3\widehat{i} + 6\widehat{j} + (\lambda - 5)\widehat{k}$$

The vector $$\overrightarrow{v} = \widehat{i} + \widehat{j} + \widehat{k}$$ is given. The projection of $$\overrightarrow{v}$$ onto $$\overrightarrow{AC}$$ has a length of 1 unit. The scalar projection of a vector $$\overrightarrow{a}$$ onto $$\overrightarrow{b}$$ is given by $$\frac{|\overrightarrow{a} \cdot \overrightarrow{b}|}{|\overrightarrow{b}|}$$, and this equals 1:

$$\left| \frac{\overrightarrow{v} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|} \right| = 1$$

$$|\overrightarrow{v} \cdot \overrightarrow{AC}| = |\overrightarrow{AC}|$$

$$\overrightarrow{v} \cdot \overrightarrow{AC} = (1)(3) + (1)(6) + (1)(\lambda - 5) = 3 + 6 + \lambda - 5 = \lambda + 4$$
$$|\overrightarrow{AC}| = \sqrt{3^2 + 6^2 + (\lambda - 5)^2} = \sqrt{9 + 36 + (\lambda - 5)^2} = \sqrt{45 + (\lambda - 5)^2}$$

$$|\lambda + 4| = \sqrt{45 + (\lambda - 5)^2}$$

$$\lambda = 3$$

Verify by substituting $$\lambda = 3$$:

$$\overrightarrow{AC} = 3\widehat{i} + 6\widehat{j} + (3 - 5)\widehat{k} = 3\widehat{i} + 6\widehat{j} - 2\widehat{k}$$

$$|\overrightarrow{AC}| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$$

$$\overrightarrow{v} \cdot \overrightarrow{AC} = 3 + 6 - 2 = 7$$

Projection length = $$\frac{|7|}{7} = 1$$, which satisfies the condition.

Now, $$\alpha > \beta$$ are the roots of the equation $$\lambda^2 x^2 - 6\lambda x + 5 = 0$$ with $$\lambda = 3$$:

$$(3)^2 x^2 - 6(3)x + 5 = 9x^2 - 18x + 5 = 0$$

$$x_1 = \frac{18 + 12}{18} = \frac{30}{18} = \frac{5}{3}$$

$$x_2 = \frac{18 - 12}{18} = \frac{6}{18} = \frac{1}{3}$$

Since $$\alpha > \beta$$, $$\alpha = \frac{5}{3}$$, $$\beta = \frac{1}{3}$$.

$$2\left(\frac{5}{3}\right) - \frac{1}{3} = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$$

Therefore, $$2\alpha - \beta = 3$$.