Concept:

Line in 3D: 

Equation of line passing through a point P(x1, y1, z1) with direction cosine l, m, n:

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{\rm{l}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{\rm{m}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{\rm{n}}}\)

Equation of line passing through two points P(x1, y1, z1) and Q(x2, y2, z2):

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{{{\rm{y}}_2} - {{\rm{y}}_1}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{{{\rm{z}}_2} - {{\rm{z}}_1}}}\)

Note:

Let direction ratios of a line be (l1, m1, n1) and for another line be (l2, m2, n2) then:

If lines are perpendicular, l1l2 + m1m2 + n1n2 = 0

If lines are parallel, \(\frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} = \frac{{{{\rm{m}}_1}}}{{{{\rm{m}}_2}}} = \frac{{{{\rm{n}}_1}}}{{{{\rm{n}}_2}}}\)

Calculation:

Given:

A = (1, 0, 3)

B = (4, 7, 1)

C = (3, 5, 3)

Direction ratio of line BC is (-1, -2, 2)

Equation of BC,

\(\frac{{{\rm{x}} - 4}}{{3 - 4}} = \frac{{{\rm{y}} - { 7} }}{{5-7}} = \frac{{{\rm{z}} - 1}}{{3-1}}\)

\(\Rightarrow \frac{{{\rm{x}} - 4}}{{-1}} = \frac{{{\rm{y}} - { 7} }}{{-2}} = \frac{{{\rm{z}} - 1}}{{2}} =\rm k\)

⇒ x = -k + 4, y = -2k + 7 and z = 2k + 1

General point on BC be D = (-k + 4, -2k + 7, 2k + 1) and let D be foot of perpendicular.

Direction ratio of line AD will be (-k + 4 - 1), (-2k + 7 - 0), (2k + 1 - 3)

⇒ (-k + 3, -2k + 7, 2k - 2) 

Now, since the line BC and AD are perpendicular, l1l2 + m1m2 + n1n2 = 0

So, (-k + 3) × -1 + (-2k + 7) × -2 + (2k - 2) × 2 = 0

⇒ k - 3 + 4k - 14 + 4k - 4 = 0

⇒ 9k = 21

⇒ k = \(\frac 73\)

So, D = \(\left( \frac 5 3 , \frac 7 3 , \frac {17}{3}\right)\)