JEE Main 2025 Apr 7 Shift 2, Mathematics Q14: Lines in Space

Since the point $\left(1,-4,c\right)$ (read from the given symmetric form) and the point $\left(0,-\frac{1}{2},0\right)$ both lie on the required line, its direction ratios can also be found from these two points, and substituting into the symmetric form directly gives:

$\frac{0-1}{-2}=\frac{-\frac{1}{2}+4}{d}=\frac{0-c}{-4}\Rightarrow \frac{1}{2}=\frac{7/2}{d}=\frac{c}{4}$, giving $c=2$ and $d=7$.

The required line is perpendicular to both given lines, so its direction is along $(\hat{i}+a\hat{j}+b\hat{k})\times (-b\hat{i}+a\hat{j}+5\hat{k})$, which works out to $\left(5a-ab\right)\hat{i}-(b^2+5)\hat{j}+(a+ab)\hat{k}$.

Matching this direction, up to a scalar, with the known direction $(-2,7,-4)$:

$\frac{5a-ab}{-2}=\frac{-(b^2+5)}{7}=\frac{a+ab}{-4}$

Equating the first and third ratios: $-4(5a-ab)=-2(a+ab)\Rightarrow -20a+4ab=-2a-2ab\Rightarrow 6a(b-3)=0\Rightarrow b=3$ (taking $a\ne 0$).

Substituting $b=3$ into the first and second ratios: $\frac{5a-3a}{-2}=\frac{-14}{7}=-2\Rightarrow \frac{2a}{-2}=-2\Rightarrow a=2$.

$\therefore a+b+c+d=2+3+2+7=14$.