The point of intersection of the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$ is :

$\left(\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\right)$

Any point on the first line is

(3r₁ – 1, 5r₁ – 3, 7r₁ – 5)

and any point on the second line is

(r₂ + 2, 3r₂ + 4, 5r₂ + 6)

At the point of intersection, we must have

3r₁ – 1 = r₂ + 2,

5r₁ – 3 = 3r₂ + 4

7r₁ – 5 = 5r₁ + 6

Thus, $r_1=\frac{1}{2}, r_2=-\frac{3}{2}$

Hence required point is $\frac{3}{2}-1, \frac{5}{2}-3, \frac{7}{2}-5$ i.e. $\left(\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\right)$