Let ${\mathrm{L}}_1:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{2}$ and ${\mathrm{L}}_2:\frac{\mathrm{x}+1}{-1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}}{|1}$ be two lines. Let L₃ be a line passing through the point ($\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$) and be perpendicular to both L₁ and L₂. If L₃ intersects L₁, then $|5\mathrm{\alpha }-11\mathrm{\beta }-8\mathrm{\gamma }|$ equals:

DR’s of ${\mathrm{L}}_3=\overrightarrow{\mathrm{m}}\times \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 1& -1& 2\\ -1& 2& 1\end{array}\right|$
$\begin{array}{l}=-5\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\\ {\mathrm{L}}_3:\frac{\mathrm{x}-\mathrm{\alpha }}{-5}=\frac{\mathrm{y}-\mathrm{\beta }}{-3}=\frac{\mathrm{z}-\mathrm{\gamma }}{1}=\mathrm{\lambda }\\ \mathrm{A}(\mathrm{\alpha }-5\mathrm{\lambda },\mathrm{\beta }-3\mathrm{\lambda },\mathrm{\gamma }+\mathrm{\lambda })\\ {\mathrm{L}}_1:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{2}=\mathrm{k}\end{array}$
B(k + 1, –k + 2, 2k + 1)
Now
$\begin{array}{l}\mathrm{\alpha }-5\mathrm{\lambda }=\mathrm{k}+1\Rightarrow \mathrm{\alpha }=5\mathrm{\lambda }+\mathrm{k}+1\\ \mathrm{\beta }-3\mathrm{\lambda }=-\mathrm{k}+2\Rightarrow \mathrm{\beta }=3\mathrm{\lambda }-\mathrm{k}+2\\ \mathrm{\gamma }+\mathrm{\lambda }=2\mathrm{k}-1\Rightarrow \mathrm{\gamma }=-\mathrm{\lambda }+2\mathrm{k}+1\\ |5\mathrm{\alpha }-11\mathrm{\beta }-8\mathrm{\gamma }|=|-25|\end{array}$
= 25