Find the vector equation of the line passing through the point $(1,2, - 4)$ and perpendicular to the two lines:
$\cfrac{{x - 8}}{3} = \cfrac{{y + 19}}{{ - 16}} = \cfrac{{z - 10}}{7}$ and $\cfrac{{x - 15}}{3} = \cfrac{{y - 29}}{8} = \cfrac{{z - 5}}{{ - 5}}$

.: Any line through $(1,2, - 4)$ is
$\vec r = (\hat i + 2\hat j - 4\hat k) + \lambda ({b_1}\hat i + {b_2}\hat j + {b_3}\hat k)$ …(1)

where, ${b_1}\hat i + {b_2}\hat j + {b_3}\hat k$ its direction
The line $\cfrac{{x - 8}}{3} = \cfrac{{y + 19}}{{ - 16}} = \cfrac{{z - 10}}{7}$ ….(2)

has direction $3\hat i - 16\hat j + 7\hat k.$

Now, lines (1) and (2) are perpendicular,

so
$({b_1}\hat i + {b_2}\hat j + {b_3}\hat k) \cdot (3\hat i - 16\hat j + 7\hat k) = 0 \Rightarrow 3{b_1} - 16{b_2} + 7{b_3} = 0$ ….(3)

Similarly $3{b_1} + 8{b_2} - 5{b_3} = 0$ ...(4)

Solving (3) and (4),

we get
$\cfrac{{{b_1}}}{{80 - 56}} = \cfrac{{{b_2}}}{{21 + 15}} = \cfrac{{{b_3}}}{{24 + 48}}$

$\Rightarrow$ $\cfrac{{{b_1}}}{{24}} = \cfrac{{{b_2}}}{{36}} = \cfrac{{{b_3}}}{{72}} \Rightarrow \cfrac{{{b_1}}}{2} = \cfrac{{{b_2}}}{3} = \cfrac{{{b_3}}}{6}$

$\therefore$ The direction of (1) is $2\hat i + 3\hat j + 6\hat k.$

$\therefore$ The equation of line (1) is $\vec r = \hat i + 2\hat j - 4\hat k + \lambda (2\hat i + 3\hat j + 6\hat k)$ .