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 (3, -1, 1) and (-3, 2, 4) respectively.

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

x_{1} = 3s+3 , y_{1} = -s+8 , z_{1} = s+3

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

x_{2} = -3t – 3 , y_{2} = 2t – 7 , z_{2} = 4t + 6

Direction ratios of are ((-3t – 3s - 6), (2t + s - 15), (4t – s + 3))

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

Therefore, by the condition of perpendicularity,

3(-3t – 3s - 6) – 1(2t + s - 15) + 1(4t – s + 3) = 0 and

-3(-3t – 3s - 6) + 2(2t + s - 15) + 4(4t – s + 3) = 0

⇒ -9t – 9s – 18 – 2t – s + 15 + 4t – s + 3 = 0 and

9t + 9s + 18 + 4t + 2s – 30 + 16t – 4s + 12 = 0

⇒ -7t – 11s = 0 and

29t + 7s = 0

Solving above two equations, we get,

t = 0 and s = 0

therefore,

P ≡ (3, 8, 3) and Q ≡ (-3, -7, 6)

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

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