Let $$\vec{a} = \alpha\hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}$$, $$\alpha > 0$$. If the projection of $$\vec{a} \times \vec{b}$$ on the vector $$-\hat{i} + 2\hat{j} - 2\hat{k}$$ is $$30$$, then $$\alpha$$ is equal to

We need to find $$\alpha$$ given that the projection of $$\vec{a} \times \vec{b}$$ on $$-\hat{i} + 2\hat{j} - 2\hat{k}$$ is 30.

First, with $$\vec{a} = \alpha\hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}$$, we compute $$\vec{a} \times \vec{b}$$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & -1 \\ 2 & 1 & -\alpha \end{vmatrix}$$

$$= \hat{i}(1 \cdot (-\alpha) - (-1) \cdot 1) - \hat{j}(\alpha \cdot (-\alpha) - (-1) \cdot 2) + \hat{k}(\alpha \cdot 1 - 1 \cdot 2)$$

$$= \hat{i}(-\alpha + 1) - \hat{j}(-\alpha^2 + 2) + \hat{k}(\alpha - 2)$$

$$= (1 - \alpha)\hat{i} + (\alpha^2 - 2)\hat{j} + (\alpha - 2)\hat{k}$$

Next, let $$\vec{c} = -\hat{i} + 2\hat{j} - 2\hat{k}$$ so that $$|\vec{c}| = \sqrt{1 + 4 + 4} = 3$$, and compute the projection of $$\vec{a} \times \vec{b}$$ on $$\vec{c}$$:

Projection = $$\dfrac{(\vec{a} \times \vec{b}) \cdot \vec{c}}{|\vec{c}|}$$

$$ (\vec{a} \times \vec{b}) \cdot \vec{c} = (1 - \alpha)(-1) + (\alpha^2 - 2)(2) + (\alpha - 2)(-2)$$

$$= -(1 - \alpha) + 2\alpha^2 - 4 - 2\alpha + 4$$

$$= -1 + \alpha + 2\alpha^2 - 4 - 2\alpha + 4$$

$$= 2\alpha^2 - \alpha - 1$$

Then substituting into the projection formula gives $$\dfrac{2\alpha^2 - \alpha - 1}{3} = 30$$, which leads to $$2\alpha^2 - \alpha - 1 = 90$$ and hence $$2\alpha^2 - \alpha - 91 = 0$$.

Finally, solving $$2\alpha^2 - \alpha - 91 = 0$$ yields $$\alpha = \dfrac{1 \pm \sqrt{1 + 728}}{4} = \dfrac{1 \pm \sqrt{729}}{4} = \dfrac{1 \pm 27}{4}$$, so $$\alpha = \dfrac{28}{4} = 7$$ or $$\alpha = \dfrac{-26}{4}$$ (rejected since $$\alpha > 0$$).

The correct answer is Option D: $$\alpha = 7$$.