Three lines are given by → r = λ ^ i , λ ∈ R → r = μ( ^ i + ^ j ),...

Finding point $‘A’$
Line \overset{\to}{r} = \lambda\overset{^}{i} and plane $x + y + z = 1$
General point of line $\left(\lambda, 0, 0\right),$ satisfy the plane
$\lambda + 0 + 0 = 1$
$\lambda = 1$
Hence, $A ( 1, 0, 0 )$ ...(i)
For point $‘B’$
General point of line \overset{\to}{r} = \mu\left(\overset{^}{i} + \overset{^}{j}\right), B\left(\mu, \mu, 0\right)
Satisfy the plane and $\mu + \mu + 0 = 1$
$\mu = \dfrac{1}{2}$
Point $B\left(\dfrac{1}{2}, \dfrac{1}{2}, 0\right)$ ...(ii)
Similarly for point $‘C’$
General point of line \overset{\to}{r} = v\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right), C ( v, v, v )
Satisfy the plane $v + v + v = 1$
$v = \dfrac{1}{3}$
Point $C\left(\dfrac{1}{3}, \dfrac{1}{3}, \dfrac{1}{3}\right)$ ....(iii)
⇒ \overset{\to}{AB} = - \dfrac{1}{2}\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} and \overset{\to}{AC} = - \dfrac{2}{3}\overset{^}{i} + \dfrac{1}{3}\overset{^}{j} + \dfrac{1}{3}\overset{^}{k}
Now area of triangle $\delta ABC = \dfrac{1}{2}\left|\overset{\to}{AB} \times \overset{\to}{AC}\right|$
\Delta = \left|\dfrac{1}{2}\left|\begin{cases}i & j & k \\ - \dfrac{1}{2} & \dfrac{1}{2} & 0 \\ - \dfrac{2}{3} & \dfrac{1}{3} & \dfrac{1}{3}\end{cases}\right|\right| = \left|\dfrac{1}{2}\left(\overset{^}{i}6 + \overset{^}{j}6 + \overset{^}{k}6\right)\right| = \dfrac{\sqrt{3}}{12}

$⇒ {\left(6 ∆\right)}^{2} = \dfrac{3}{4} = 0 . 75$