**Question: **Find the distance between the skew lines L

_{1} and L

_{2} whose parametric equations are given as follows:

L

_{1} : x = 1 + 2t, y = 3t and z = 2 - t.

L

_{2} : x = -1 + s, y = 4 + s, z = 1 + 3s.

** Solution: **

**Step 1: Finding equation of a plane containing the line L**_{2}.

Let P_{1} and P_{2} be the two parallel planes containing L_{1} and L_{2}. The common normal vector of these planes 'n' must be orthogonal to both v_{1} = <2, 3, -1> (direction of L_{1}) and v_{2} = <1, 1, 3> (direction of L_{2}).

The normal vector n = v_{1} x v_{2} = $\begin{vmatrix}

i &j &k\\

2& 3 &-1 \\

1 &1 &3

\end{vmatrix}$ = i(9 + 1) -j(6 + 1) +k(2 - 3) = 10i -7j - k.

If we put s =0, in the equation of L_{2}, we get a point on L_{2} as (-1, 4, 1).

Thus the equation of the plane P_{2} is,

a(x - x_{0}) + B(y - y_{0}) + c(z -z_{0}) = 10

10(x + 1) -7(y - 4) -(z -1) = 0

10x - 7y - z + 39 = 0 Equation of plane P_{2}.

**Step 2: Finding a point on line L**_{1}.

If we let t = 0 in the equation of L_{1}, we get the point on L_{1} as (1, 0, 2).

**Step 3: Use distance formula to find the distance between the skew lines**

Now we need to find the distance between the point (1, 0, 2) and the plane P_{2}, using the formula

D = $\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$

= $\frac{|10(1)-7(0)-1(2)+39|}{\sqrt{10^{2}+(-7)^{2}+(-1)^{2}}}$

= $\frac{47}{\sqrt{150}}$ ≈ 3.84 units

The distance between the skew lines L_{1} and L_{2} ≈ 3.84 units.