उत्तर: 18

समाधान:

सूत्र: सीधी रेखा का समीकरण , असंगत रेखाएँ

$ \begin{aligned} & \overrightarrow{{}r}=(i+2 j+3 k)+\lambda(i+j+k) \Rightarrow \overrightarrow{{}r}=\overrightarrow{{}a}+\lambda \overrightarrow{{}p} \\ & \overrightarrow{{}r}=(+\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}-\hat{j}) \Rightarrow \overrightarrow{{}r}=\overrightarrow{{}b}+\mu \overrightarrow{{}q} \\ & \overrightarrow{{}p} \times \overrightarrow{{}q}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} =\hat{i}+2 \hat{j}-3 \hat{k} \\ & d=|\dfrac{(\vec{b}-\vec{a}) \cdot(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}| \\ & d=|\dfrac{(-3 \hat{j}-\hat{k}) \cdot(\hat{i}+2 \hat{j}-3 \hat{k})}{\sqrt{14}}| \\ & =|\dfrac{-6+3}{\sqrt{14}}|=\dfrac{3}{\sqrt{14}} \\ & \alpha=\dfrac{3}{\sqrt{14}} \end{aligned} $

अब, $28 \alpha^{2}= 28 \times \dfrac{9}{14}=18$