If the equation of the line passing through the point (0,-(1)/(2),...

Line is $\perp^{\mathrm{r}}$ to 2 line $\Rightarrow$ line will be parallel to$(i+a \hat{j}+b \hat{k}) \times(-b \hat{i}+a \hat{j}+5 \hat{k})$Parallel vector along the required line is$\hat{\mathrm{i}}(5 \mathrm{a}-\mathrm{ab})-\hat{\mathrm{j}}\left(\mathrm{ b}^2+5\right)+\hat{\mathrm{k}}(\mathrm{a}+\mathrm{ab})$Dr's of required line $\alpha(5 a-a b),-\left(b^2+5\right),(a+a b)Also Dr's of required line$\alpha-2, $\mathrm{ d},-$4Also point \left(0, \frac{-1}{2}, 0\right)$will lie on$\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{ d}}=\frac{\mathrm{z}-\mathrm{c}}{-4} \frac{0-1}{-2}=\frac{\frac{-1}{2}+4}{d}=\frac{0-c}{-4} \Rightarrow d=7, c=2 $$\begin{aligned}&\text{ From (1) } \frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{-\mathrm{b}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\\ &\frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{\mathrm{a}+\mathrm{ab}}{-4} ; \frac{-\mathrm{b}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\end{aligned}$$ $$\begin{array}{c|c}-20 a+4 a b=-2 a-2 a b & 4 b^2+20=70+7 a b \\18 a=6 a b & 36+20=70+21 a \b=3 & 56=28 a \Rightarrow a=2\end{array}$$ a+b+c+d=2+3+2+7=14