Let P (-2,-1,1) and Q left(frac{56}{17}, frac | vedclass

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(C) In a rhombus $PRQS$,the diagonals $PQ$ and $RS$ are perpendicular to each other.The direction ratios of diagonal $PQ$ are given by $(x_Q - x_P, y_

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$450$

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(C) In a rhombus $PRQS$,the diagonals $PQ$ and $RS$ are perpendicular to each other.The direction ratios of diagonal $PQ$ are given by $(x_Q - x_P, y_Q - y_P, z_Q - z_P) = \left(\frac{56}{17} + 2, \frac{43}{17} + 1, \frac{111}{17} - 1\right) = \left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}\right)$.Since $PQ \perp RS$,the dot product of their direction ratios must be zero.Let the direction ratios of $RS$ be $(\alpha, -1, \beta)$. Then,$\frac{90}{17}(\alpha) + \frac{60}{17}(-1) + \frac{94}{17}(\beta) = 0$.Multiplying by $17$,we get $90\alpha - 60 + 94\beta = 0$,which simplifies to $90\alpha + 94\beta = 60$,or $45\alpha + 47\beta = 30$.We need to find integers $\alpha$ and $\beta$ with minimum absolute values satisfying $47\beta = 30 - 45\alpha$.Testing values: If $\alpha = -15$,then $47\beta = 30 - 45(-15) = 30 + 675 = 705$. Thus,$\beta = \frac{705}{47} = 15$.Wait,checking the equation $45\alpha + 47\beta = 30$: If $\alpha = -15$,$45(-15) + 47\beta = 30 \Rightarrow -675 + 47\beta = 30 \Rightarrow 47\beta = 705 \Rightarrow \beta = 15$.Thus,$\alpha^2 + \beta^2 = (-15)^2 + 15^2 = 225 + 225 = 450$.

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