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

Explanation

Equation of a line passing through the point `(1,2,-4)`
`(x-1)/(a) = (y-2)/(b) = (z+4)/(c) "……"(1)`
where a, b and c are the direction ratio of line (1).
Given lines are
`(x-8)/(3) = (y+19)/(-16) = (z+4)/(7)`
and `(x-15)/(3)= (y-29)/(8) = (z-5)/(-5)`
The direction ratios of the above line are `(3,-16,7)` and (3,8,-5) respectively and these are perpendicular to lines (1),
`:. 3a-16b+7c=0"......"(2)`
and `3a+8b-5c=0"......"(3)`
From cross multiplication method,
`(a)/(80-56)= (b)/(21+15) = (c )/(24+48)`
`rArr a/24=(b)/(26) = (c)/(72) rArr (a)/(2) = (b)/(3) = (c )/(6) = lambda`(say)
`:. a = 2 lambda , b = 3lambda` and `c = 6lambda`
Therefore, equation of a line through the point `(1,2,-4)` and parallel to vector `2hati+3hatj+6hatk` is `vecr = (hati+2hatj-4hatk) + lambda(2hati+3hatj+6hatk)`.