Let $$P(-2, -1, 1)$$ and $$Q\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $$\alpha, -1, \beta$$, where both $$\alpha$$ and $$\beta$$ are integers of minimum absolute values, then $$\alpha^2 + \beta^2$$ is equal to

We have a rhombus $$PRQS$$ with vertices $$P(-2, -1, 1)$$ and $$Q\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$$.

In a rhombus, the diagonals bisect each other. So the midpoint of $$PQ$$ is also the midpoint of $$RS$$.

Midpoint $$= \left(\frac{-2 + 56/17}{2}, \frac{-1 + 43/17}{2}, \frac{1 + 111/17}{2}\right)$$

$$= \left(\frac{-34/17 + 56/17}{2}, \frac{-17/17 + 43/17}{2}, \frac{17/17 + 111/17}{2}\right)$$

$$= \left(\frac{22/17}{2}, \frac{26/17}{2}, \frac{128/17}{2}\right) = \left(\frac{11}{17}, \frac{13}{17}, \frac{64}{17}\right)$$

$$\vec{PQ} = \left(\frac{56}{17}+2, \frac{43}{17}+1, \frac{111}{17}-1\right) = \left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}\right)$$

Direction ratios of $$PQ$$: $$(90, 60, 94)$$, or simplified $$(45, 30, 47)$$.

In a rhombus, the diagonals are perpendicular. So $$\vec{RS} \perp \vec{PQ}$$.

Direction ratios of RS are given as $$(\alpha, -1, \beta)$$.

$$\vec{RS} \cdot \vec{PQ} = 0$$:

$$45\alpha + 30(-1) + 47\beta = 0$$

$$45\alpha + 47\beta = 30$$

We need integers $$\alpha, \beta$$ satisfying $$45\alpha + 47\beta = 30$$ with minimum $$|\alpha|$$ and $$|\beta|$$.

By inspection or extended Euclidean algorithm:

$$\gcd(45, 47) = 1$$, so solutions exist.

$$47 = 1 \times 45 + 2$$

$$45 = 22 \times 2 + 1$$

Back-substituting: $$1 = 45 - 22 \times 2 = 45 - 22(47 - 45) = 23 \times 45 - 22 \times 47$$

So $$30 = 30 \times 23 \times 45 - 30 \times 22 \times 47 = 690 \times 45 - 660 \times 47$$.

General solution: $$\alpha = 690 - 47k$$, $$\beta = -660 + 45k$$.

To minimize $$|\alpha|$$: $$690 - 47k \approx 0 \Rightarrow k \approx 14.68$$.

At $$k = 15$$: $$\alpha = 690 - 705 = -15$$, $$\beta = -660 + 675 = 15$$.

At $$k = 14$$: $$\alpha = 690 - 658 = 32$$, $$\beta = -660 + 630 = -30$$.

Check $$k = 15$$: $$45(-15) + 47(15) = -675 + 705 = 30$$. Correct.

For minimum absolute values, we want $$|\alpha|$$ and $$|\beta|$$ both as small as possible.

At $$k = 15$$: $$(\alpha, \beta) = (-15, 15)$$, $$\alpha^2+\beta^2 = 225+225 = 450$$.

At $$k = 14$$: $$(\alpha, \beta) = (32, -30)$$, $$\alpha^2+\beta^2 = 1024+900 = 1924$$.

At $$k = 16$$: $$(\alpha, \beta) = (-62, 60)$$, $$\alpha^2+\beta^2 = 3844+3600 = 7444$$.

The minimum is at $$k = 15$$: $$\alpha = -15$$, $$\beta = 15$$.

$$\alpha^2 + \beta^2 = 225 + 225 = 450$$.

The answer is 450.