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

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

x_{1} = -s+3 , y_{1} = 2s+4 , z_{1} = s-2

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

x_{2} = t+1 , y_{2} = 3t – 7 , z_{2} = 2t - 2

Direction ratios of are ((t + s – 2), (3t – 2s – 11), (2t – s))

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

Therefore, by the condition of perpendicularity,

-1(t + s – 2) + 2(3t – 2s – 11) + 1(2t – s) = 0 and

1(t + s – 2) + 3(3t – 2s – 11) + 2(2t – s) = 0

⇒ - t – s + 2 + 6t – 4s – 22 + 2t – s = 0 and

t + s – 2 + 9t – 6s – 33 + 4t – 2s = 0

⇒ 7t – 6s = 20 and

14t - 7s = 35

Solving above two equations, we get,

t = 2 and s = -1

therefore,

P ≡ (4, 2, -3) and Q ≡ (3, -1, 2)

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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