Given that line passing through the point (−1, 3, 5) and parallel to line  \(\cfrac{x-3}2\) = \(\cfrac{y-3}3\) =\(\cfrac{z-2}1\). Now, position vector of point (−1, 3, 5) is \(\vec a=-\hat i+3\hat j+5\hat k \).

And direction ratios of given line  \(\cfrac{x-3}2\) = \(\cfrac{y-3}3\) =\(\cfrac{z-2}1\). are 2, 3 and 1.

Now, equation of line parallel to given line is \(\vec b=2\hat i+3\hat j+\hat k \).

We know that vector form of line passing through the point p and parallel to vector \(\vec b\) is given by  \(\vec r=\vec a+\lambda\vec b\) where  \(\vec a\) is position vector of point p.

Therefore, the vector equation of line passing through the point (−1, 3, 5) and parallel to vector \(\vec b\) is  \(\vec r=\vec a+\lambda\vec b\) = \((-\hat i+3\hat j+5\hat k) \) + \(\lambda(2\hat i+3\hat j+\hat k) \).

(Because  \(\vec b\) is parallel to given line and we want to find the line parallel to given line. Therefore, required line is parallel to \(\vec b\). )

Hence, the vector equation of required line is  \(\vec r=\)  \((-\hat i+3\hat j+5\hat k) \) + \(\lambda(2\hat i+3\hat j+\hat k) \).