JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Three Dimensional Geometry

With direction vectors ${\vec{b}}_1=(2,3,4)$ and ${\vec{b}}_2=(1,\alpha ,1)$ through the points $(1,2,3)$ and $(0,0,5)$, the cross product is ${\vec{b}}_1\times {\vec{b}}_2=(3-4\alpha ,2,2\alpha -3)$. The joining vector is $(-1,-2,2)$, so the numerator of the shortest-distance formula is $|-(3-4\alpha )-2(2)+2(2\alpha -3)|=|8\alpha -13|$, while $|{\vec{b}}_1\times {\vec{b}}_2{|}^2=(3-4\alpha {)}^2+4+(2\alpha -3{)}^2=20{\alpha }^2-36\alpha +22$. Setting $\frac{|8\alpha -13|}{\sqrt{20{\alpha }^2-36\alpha +22}}=\frac{5}{\sqrt{6}}$ and squaring gives $6(8\alpha -13{)}^2=25(20{\alpha }^2-36\alpha +22)$, which simplifies to ${\alpha }^2+3\alpha -4=0$. The roots $\alpha =1$ and $\alpha =-4$ sum to $-3$.