Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

(A) The equations of the given lines are $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$.
The direction ratios of the first line are $a_{1}=7, b_{1}=-5, c_{1}=1$.
The direction ratios of the second line are $a_{2}=1, b_{2}=2, c_{2}=3$.
Two lines with direction ratios $(a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ are perpendicular to each other if and only if $a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0$.
Substituting the values,we get:
$a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (7 \times 1) + (-5 \times 2) + (1 \times 3)$
$= 7 - 10 + 3$
$= 0$.
Since the sum of the products of the corresponding direction ratios is $0$,the given lines are perpendicular to each other.