Given: A(1, 8, 4)

Line segment joining B(0, -1, 3) and C(2, -3, -1) is

BC = 2i – 2j – 4k

Let the foot of the perpendicular be R then,

As R lies on the line having point B and parallel to BC,

So, R = (0, -1, 3) + a(2, -2, -4)

R(2a, -1-2a, 3-4a)

The line segment AR is

AR = (2a-1)i + (-1-2a-8)j + (3-4x-4)k

As the lines AR and BC are perpendicular thus, (as R is the foot of the perpendicular on BC)

AR.BC = 0

ð 2(2a-1) + (-2)(-9-2a) + (-4)(-1-4a) = 0

ð 24a + 20 = 0

ð a =

Substituting a in R we get,

R()