Let P - 2 , - 1 , 1 and Q 56 17 , 43 17 , 111 17 be the vertices of the rhombus P R Q S . If the direction…

Given,

 $P\left(-2,-1,1\right)$ 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

Now $RS=\left(\alpha ,-1,\beta \right)$

So, direction ratio of $PQ\equiv \left(\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1\right)$

$\equiv \left(\frac{90}{17},\frac{60}{17},\frac{94}{17}\right)$

Now we know that diagonal of rhombus are perpendicular,

So, $\frac{90}{17}\alpha +\frac{60}{17}\left(-1\right)+\frac{94}{17}\beta =0$

$\Rightarrow 90\alpha +94\beta =60$

$\Rightarrow \beta =\frac{60-90\alpha }{94}$

$\Rightarrow \beta =\frac{30\left(2-3\alpha \right)}{94}$

$\Rightarrow \beta =-30\frac{\left(3\alpha -2\right)}{94}$

$\Rightarrow \beta =\frac{-15}{47}\left(3\alpha -2\right)$

$\Rightarrow \frac{\beta }{-15}=\frac{3\alpha -2}{47}$

$\Rightarrow \beta =-15,\alpha =-15$ $\left\{\text{as }\alpha \text{ \& }\beta \text{ are integer}\right\}$

So, ${\alpha }^2+{\beta }^2=225+225$$=450$