Any line through (1, 2, –4) is $\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})$ $+\lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$ ...(1) The line $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ ...(2) has direction $3\hat{i}-16\hat{j}+7\hat{k}$ Now lines (1) and (2) are perpendicular, we get $(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})\cdot(3\hat{i}-16\hat{j}+7\hat{k})=0$$\Rightarrow 3b_1 -16b_2 + 7b_3 = 0$...(3)Similarly $3b_1 + 8b_2 - 5b_3 = 0$ ...(4)Solving (3) and (4), we get$\frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}$∴ The direction of (1) is $2\hat{i}+3\hat{j}+6\hat{k}$∴ The equation of line (1) is $\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})$ $+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$