JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Three Dimensional Geometry

The required line is perpendicular to both given lines, so its direction is along $(\hat{i}+a\hat{j}+b\hat{k})\times (-b\hat{i}+a\hat{j}+5\hat{k})=(5a-ab)\hat{i}-(b^2+5)\hat{j}+(a+ab)\hat{k}$, and this is proportional to $(-2,d,-4)$. Since the line passes through $\left(0,-\frac{1}{2},0\right)$, substituting into $\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}$ gives the common value $\frac{1}{2}$, so $\frac{7/2}{d}=\frac{1}{2}$ and $\frac{-c}{-4}=\frac{1}{2}$, i.e. $d=7$ and $c=2$. Matching the direction ratios, $\frac{5a-ab}{-2}=\frac{a+ab}{-4}$ gives $b=3$, and then $\frac{-(b^2+5)}{7}=\frac{a+ab}{-4}$ gives $a=2$. Hence $a+b+c+d=2+3+2+7=14$.