Solution:

It is given that
The Cartesian equation is
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \ldots \text { (1) }$
It is known to us that
The Cartesian eq. of a line passing through a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and having direction cosines $1, \mathrm{~m}, \mathrm{n}$ is
$\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$
On comparing this standard form with the given equation, we obtain
$\begin{array}{l}
\mathrm{x}_{1}=5, \mathrm{y}_{1}=-4, \mathrm{z}_{1}=6 \text { and } \\
1=3, \mathrm{~m}=7, \mathrm{n}=2
\end{array}$
The point through which the line passes has the position vector $\vec{a}=5 \hat{i}-4 \hat{j}+6 k$ and
The vector parallel to the line is given by $\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}$
As, vector eq. of a line that passes through a given point whose position vector is $\vec{a}$ and parallel to a given vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$
As a result, the required line in vector form is given as:
$\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$