The given cartesian equation (in symmetrical form) is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$⇒ The line passes through the point (5, – 4, 6) and is parallel to vector $3\hat{i}+7\hat{j}+2\hat{k}.$ Hence, the vector equation of the line is $\vec{r}= (5\hat{i}-4\hat{j}+6\hat{k})$ $+\lambda(3\hat{i}+7\hat{j}+2\hat{k}).$