The given equation of line is $\frac{\mathit{x}-5}{3}$ = $\frac{\mathit{y}+4}{7}$ = $\frac{6-\mathit{z}}{2}$
i.e in standard form $\frac{\mathit{x}-5}{3}$ = $\frac{\mathit{y}-\left(-4\right)}{7}$ = $\frac{\mathit{z}-6}{-2}$
Comparing this equation with standard form $\frac{\mathit{x}-{\mathit{x}}_1}{\mathit{a}}$ = $\frac{\mathit{y}-{\mathit{y}}_1}{\mathit{b}}$ = $\frac{\mathit{z}-{\mathit{z}}_1}{\mathit{c}}$
We get, ${\mathit{x}}_1$ = 5 , ${\mathit{y}}_1$ = - 4 , ${\mathit{z}}_1$ = 6 , a = 3 , b = 7 , c = - 2
Thus, the required line is parallel to the vector $3\hat{\mathit{i}}+7\hat{\mathit{j}}-2\hat{\mathit{k}}$ and passes through the point (5, -4, 6).
The vector form of the line can be written as $\overrightarrow{\mathit{r}}$ = $\overrightarrow{\mathit{a}}+\mathit{\lambda }\overrightarrow{\mathit{b}}$ , where λ is a constant
Thus, the required equation is $\overrightarrow{\mathit{r}}$ = $\left(5\hat{\mathit{i}}-4\hat{\mathit{j}}+6\hat{\mathit{k}}\right)$ + λ $\left(3\hat{\mathit{i}}+7\hat{\mathit{j}}-2\hat{\mathit{k}}\right)$