The given equation of line is $\frac{x-5}{3}$ = $\frac{y+4}{7}$ = $\frac{6-z}{2}$ i.e in standard form $\frac{x-5}{3}$ = $\frac{y-(-4)}{7}$ = $\frac{z-6}{-2}$ Comparing this equation with standard form $\frac{x-x_1}{a}$ = $\frac{y-y_1}{b}$ = $\frac{z-z_1}{c}$ We get, $x_1$ = 5 , $y_1$ = - 4 , $z_1$ = 6 , a = 3 , b = 7 , c = - 2 Thus, the required line is parallel to the vector $3\hat{i}+7\hat{j}-2\hat{k}$ and passes through the point (5, -4, 6). The vector form of the line can be written as $\vec{r}$ = $\vec{a}+\lambda\vec{b}$ , where λ is a constant Thus, the required equation is $\vec{r}$ = $(5\hat{i}-4\hat{j}+6\hat{k})$ + λ $(3\hat{i}+7\hat{j}-2\hat{k})$