Given lines are \(\frac {x-1}{2} = \frac {2-y}{3} = \frac {z+1}{4}\)

= \(\frac {x-1}{2} = \frac {y-2}{-3} = \frac {z-(-1)}{4}...(1)\)

∴ x₁ = 1, y₁ = 2, z₁ = -1  and a₁ = 2, b₁ = -3, c₁ = 4

And \(\frac {x+2}{-1} =\frac {y-3}{2} = \frac z3\)

= \(\frac {x-(-2)}{-1} = \frac {y-3}{2} = \frac {z-0}{3}\).....(2)

∴ x₂ = -2, y₂ = 3, z₂ = 0 and a₂ = -1, b₂ = 2, c₂ = 3

Now, \(\begin {vmatrix} x_2-x_1&y_2-y_1&z_2-z_1\\ a_1 &b_1& c_1\\a_2&b_2&c_2\end{vmatrix}\)= \(\begin {vmatrix} -3&1&1\\2&-3&4\\-1&2&3\end{vmatrix}\)

= -3 \(\begin {vmatrix} -3&4\\2&3\end{vmatrix}\)-1 \(\begin {vmatrix} 2&-1\\-1&3\end{vmatrix}\)+1 \(\begin {vmatrix} 2&-3\\-1&2\end{vmatrix}\)

= -3 (9-8)-1 (6 +4) + 1 (4 -3)

= 51 - 10 + 1

= 42

(b₁ c₂-b₂ c₁)² + (c₁a₂-c₁a₁)² + (a₁ b₂ - a₂ b₁)²

= (-3 x 3 -2 x 4)² + (4 x -1 -3 x 2)² + (2 x 2 - (-1) x -3)

= (-9 -8)² + (-4-6)² + (4-3)²

= 17² + 10² + 1²

= 289 + 100 + 1

= 390

∴ Distance between both skew lines (1) & (2) is 

d= \(\frac {\begin {vmatrix} x_2-x_1&y_2-y_1&z_2-z_1\\ a_1 &b_1& c_1\\a_2&b_2&c_2\end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1 b_2-a_2b_1)^2}}\)

= \(\frac {43}{\sqrt{390}} = \frac {42}{19.75} \)= 2.1265 unit (approx)