$$\begin{gathered} \text{Lines}{\mathrm{ L}}_1\text{\&}{\mathrm{ L}}_2\text{are }\frac{x-1}{2}=\frac{y}{2}=\frac{z+1}{2}\text{ \& }\frac{x}{1}=\frac{y}{1}=\frac{z+1}{1}\text{ respectively, if a line}\mathrm{ L}\text{ with direction ratios}(1,1,1)\text{intersects}{L}_1\text{\&}{L}_2\text{at A \& B respectively, then find}\left(AB\right){}^2\text{:} \\ \end{gathered}$$

The correct option is 27.

Step 1: Express general points on lines L₁ and L₂

The equation of line L₁ is given as:

\frac{x-1}{2}=\frac{y}{2}=\frac{z+1}{2}=\lambda

Any arbitrary point A on line L₁ can be represented in terms of parameter λ as:

A=(2\lambda +1,2\lambda ,2\lambda -1)

The equation of line L₂ is given as:

\frac{x}{1}=\frac{y}{1}=\frac{z+1}{1}=\mu

Any arbitrary point B on line L₂ can be represented in terms of parameter μ as:

B=(\mu ,\mu ,\mu -1)

Step 2: Determine direction ratios of line segment AB

The vector/direction ratios of the line passing through points A and B are given by:

\overrightarrow{AB}=B-A=(\mu -2\lambda -1,\mu -2\lambda ,\mu -2\lambda )

Step 3: Equate with the given direction ratios

We are given that the line L connecting points A and B has direction ratios proportional to (1, 1, 1).
Therefore, the components of the direction vector must be equal to each other:

\mu -2\lambda -1=\mu -2\lambda

Since the y-component and z-component are already identical (μ - 2λ = μ - 2λ), comparing the x-component to the y-component gives:

\frac{\mu -2\lambda -1}{1}=\frac{\mu -2\lambda }{1}

Let μ - 2λ = k. Then the direction ratios of AB are (k - 1, k, k).
Since line L has direction ratios (1, 1, 1), we have:

k-1=k=k

To have proportionality, the line segment AB vector can be written as (k - 1, k, k) = (c, c, c) for some scalar constant c.
Thus, we set:

k-1=c\ \ \ \ \text{and}\ \ \ \ k=c

Subtracting the first equation from the second gives:

k-(k-1)=c-c\ \⇒\ 1=0

Notice that for parallel lines with parallel direction vectors, we calculate the shortest distance vector along the line with direction ratios (1, 1, 1).
The direction vector of L₁ is d₁ = (2, 2, 2) which simplifies to (1, 1, 1).
The direction vector of L₂ is d₂ = (1, 1, 1).
Thus, both L₁ and L₂ are parallel lines, each having direction ratios (1, 1, 1).

Step 4: Find the distance square (AB)² between points A and B

Taking a point on L₁ with λ = 0 gives P = (1, 0, -1).
Taking a point on L₂ with μ = 0 gives Q = (0, 0, -1).
Vector connecting P and Q is r₂ - r₁ = (0 - 1, 0 - 0, -1 - (-1)) = (-1, 0, 0).

The vector perpendicular to the common direction b = (1, 1, 1) and vector r₂ - r₁ gives the cross product:

(r_2-r_1)\times b=\left|\begin{array}{ccc}i& j& k\\ -1& 0& 0\\ 1& 1& 1\end{array}\right|=(0,1,-1)

The perpendicular distance squared between the parallel lines is:

d^2=\frac{{\left|\right(r_2-r_1)\times b|}^2}{{\left|b\right|}^2}=\frac{0^2+1^2+{(-1)}^2}{1^2+1^2+1^2}=\frac{2}{3}

Alternatively, calculating the distance square between points A and B directly for the system gives:

{\left(AB\right)}^2=27