Three lines are given by \[\overrightarrow r = \lambda \hat i\],$\lambda \in R$ , $\overrightarrow r = \mu \left( {\hat i + \hat j} \right)$ , $\mu \in R$ and $\overrightarrow r = \nu \left( {\hat i + \hat j + \hat k} \right)$ , $\nu \in R$ . Let the lines cut the plane $x + y + z = 1$ at the points $A$, $B$ and $C$ respectively. If area of the triangle $ABC$ is $\Delta $ then the value of ${\left( {6\Delta } \right)^2}$ is equal to
(A) $0.75$
(B) $0.1$
(C) $0.85$
(D) $0.65$

Hint: Here, the equations of three lines are given and also the equation of the plane is given so firstly we have to find the point of intersection of these lines and name then as $A$, $B$ and $C$ respectively. After that we can find the area of the triangle either by vector method or heron’s formula. Then after substitute the value of the area in place of $\Delta $ to get the value of ${\left( {6\Delta } \right)^2}$.

Complete step-by-step answer:
Here, the given three lines are \[\overrightarrow r = \lambda \hat i\], $\lambda \in R$ , $\overrightarrow r = \mu \left( {\hat i + \hat j} \right)$ , $\mu \in R$ and $\overrightarrow r = \nu \left( {\hat i + \hat j + \hat k} \right)$ , $\nu \in R$.
The equation of the plane is$x + y + z = 1$.
Now, the equation of first line can be written as $\overrightarrow r = \left( {\lambda \hat i + 0\hat j + 0\hat k} \right)$
Putting $\left( {\lambda ,0,0} \right)$ in the given equation of plane gives the point of intersection $A$ as we have supposed that the first line intersect the plane at $\left( {\lambda ,0,0} \right)$ .
So, $x + y + z = 1$
$ \Rightarrow \lambda + 0 + 0 = 1$
$\therefore \lambda = 1$
Point of intersection $A = \left( {1,0,0} \right)$
Now, the equation of second line can be written as $\overrightarrow r = \left( {\mu \hat i + \mu \hat j + 0\hat k} \right)$
Putting $\left( {\mu ,\mu ,0} \right)$ in the given equation of plane gives the point of intersection $B$ as we have supposed that the second line intersect the plane at $\left( {\mu ,\mu ,0} \right)$ .
So, $x + y + z = 1$
$ \Rightarrow \mu + \mu + 0 = 1$
$
   \Rightarrow 2\mu = 1 \\
  \therefore \mu = \dfrac{1}{2} \\
 $
Point of intersection $B = \left( {\dfrac{1}{2},\dfrac{1}{2},0} \right)$
Now, the equation of third line can be written as $\overrightarrow r = \left( {\nu \hat i + \nu \hat j + \nu \hat k} \right)$
Putting $\left( {\nu ,\nu ,\nu } \right)$ in the given equation of plane gives the point of intersection $C$ as we have supposed that the second line intersect the plane at $\left( {\nu ,\nu ,\nu } \right)$ .
So, $x + y + z = 1$
$ \Rightarrow \nu + \nu + \nu = 1$
$
   \Rightarrow 3\nu = 1 \\
  \therefore \nu = \dfrac{1}{3} \\
 $
Point of intersection $C = \left( {\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}} \right)$
Since the coordinates of all the three points is known so we can find the vector for the sides of triangle and then we can apply the formula for area of triangle in vector form.
Now, vector for side $AB$is given by $\left( {\dfrac{1}{2} - 1} \right)\hat i + \left( {\dfrac{1}{2} - 0} \right)\hat j + \left( {0 - 0} \right)\hat k$
$\overrightarrow {AB} = - \dfrac{1}{2}\hat i + \dfrac{1}{2}\hat j + 0\hat k$
Similarly, vector for side $AC$ is given by $\left( {\dfrac{1}{3} - 1} \right)\hat i + \left( {\dfrac{1}{3} - 0} \right)\hat j + \left( {\dfrac{1}{3} - 0} \right)\hat k$
$\overrightarrow {AC} = - \dfrac{2}{3}\hat i + \dfrac{1}{3}\hat j + \dfrac{1}{3}\hat k$
Area of the triangle( $\Delta $ ) is $ = \dfrac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {Ac} } \right|$
Now, $\Delta = \left| {\left( { - \dfrac{1}{2}\hat i + \dfrac{1}{2}\hat j} \right) \times \left( { - \dfrac{2}{3}\hat i + \dfrac{1}{3}\hat j + \dfrac{1}{3}\hat k} \right)} \right|$
By cross multiplication of two vectors we can write
$ \Rightarrow \Delta = \dfrac{1}{2}\left| {\left( {\dfrac{1}{6}} \right)\hat i + \left( {\dfrac{1}{6}} \right)\hat j + \left( {\dfrac{1}{6}} \right)\hat k} \right|$
By solving we can write
$ \Rightarrow \Delta = \dfrac{1}{2} \times \dfrac{{\sqrt 3 }}{6}$
$\therefore \Delta = \dfrac{{\sqrt 3 }}{{12}}$
So, the area of triangle $ABC$ is $\dfrac{{\sqrt 3 }}{{12}}$
Now, we have to find the value of ${\left( {6\Delta } \right)^2}$.
${\left( {6\Delta } \right)^2} = 36{\Delta ^2} = 36 \times \dfrac{3}{{144}} = \dfrac{3}{4} = 0.75$

Hence, option (A) is correct.

Note: We can also find area by heron’s formula. We know the coordinates of all vertices of the triangle so we can find the length of each side by distance formula then apply the heron’s formula.
$\Delta = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ where the $s$ is semi perimeter of the triangle and $a,b,c$ are the length of the three sides of triangle.