Answer :

__Given__**:** Cartesian equations of lines

__Formulae__**:**

**1. Condition for perpendicularity :**

If line L1 has direction ratios (a_{1}, a_{2}, a_{3}) and that of line L2 are (b_{1}, b_{2}, b_{3}) then lines L1 and L2 will be perpendicular to each other if

**2. Distance formula :**

Distance between two points A≡(a_{1}, a_{2}, a_{3}) and B≡(b_{1}, b_{2}, b_{3}) is given by,

**3. Equation of line :**

Equation of line passing through points A≡(x_{1}, y_{1}, z_{1}) and B≡(x_{2}, y_{2}, z_{2}) is given by,

__Answer__**:**

Given equations of lines

Direction ratios of L1 and L2 are (2, 1, -3) and (2, -7, 5) respectively.

Let, general point on line L1 is P≡(x_{1}, y_{1}, z_{1})

x_{1} = 2s-1 , y_{1} = s+1 , z_{1} = -3s+9

and let, general point on line L2 is Q≡(x_{2}, y_{2}, z_{2})

x_{2} = 2t+3 , y_{2} = -7t – 15 , z_{2} = 5t + 9

Direction ratios of are ((5t - 2s + 10), (-7t – s – 16), (5t + 3s))

PQ will be the shortest distance if it perpendicular to both the given lines

Therefore, by the condition of perpendicularity,

2(5t - 2s + 10) + 1(-7t – s – 16) - 3(5t + 3s) = 0 and

2(5t - 2s + 10) – 7(-7t – s – 16) + 5(5t + 3s) = 0

⇒ 10t – 4s + 20 - 7t – s - 16 - 15t – 9s = 0 and

10t - 4s + 20 + 49t + 7s + 112 + 25t + 15s = 0

⇒ -12t – 14s = -4 and

84t + 18s = -132

Solving above two equations, we get,

t = -2 and s = 2

therefore,

P ≡ (3, 3, 3) and Q ≡ (-1, -1, -1)

Now, distance between points P and Q is

Therefore, the shortest distance between two given lines is

Now, equation of line passing through points P and Q is,

⇒

Therefore, equation of line of shortest distance between two given lines is

x = y = z

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