Question

find shortest distance between the line :
(x - 8)/ 3 = (y + 19)/ -16 =( z - 10)/ 7
and
(x - 15)/ 3 =( y - 29 )/8 =( z - 5 )/ -5

Answer 1

confirm your problem...lot of = signs

Answer 2

my question is correct

Answer 3

express the lines in parametric method

$$\frac{x-8}{3}=\frac{y+19}{9}=\frac{z-10}{7}=t$$
$$x=3t+8,\quad y=9t-19,\quad z=7t+10$$
there for the firs line can be expressed in vectors as,
$$r_1(t)=\langle (3t+8),(9t-19),(7t+10)\rangle=\langle3,9,7\rangle t+\langle8,-19,10\rangle$$
Similarly, get the vector equation of the second line. It will be
$$r_2(u)=\langle 3,8,-5 \rangle u+ \langle 15,29,5 \rangle$$
The vector perpendicular to both these vectors (lines) is,
$$\mathbf{n}=\langle 3,9,7\rangle \times \langle 3,8,-5 \rangle$$

take two points on these two lines. For easiness I'll take the given points,
$$\mathbf{v}=<15,29,5>,\;\mathbf{w}=<3,8,-5>$$
v is on r_1 and w is on r_2

In theory the shortest distance S is,

$$S=\frac{\mathbf{n}}{|\mathbf{n}|}\cdot (\mathbf{v}-\mathbf{w})$$