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

Finding point $\mathrm{‘}\mathrm{A}\mathrm{’}$
Line $\overrightarrow{r}=\lambda \hat{i}$ 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 $\mathrm{‘}\mathrm{B}\mathrm{’}$
General point of line $\overrightarrow{r}=\mu \left(\hat{i}+\hat{j}\right), \mathrm{ }B\left(\mu ,\mu ,0\right)$
Satisfy the plane and $\mu +\mu +0=1$
$\mu =\frac{1}{2}$
Point $B\left(\frac{1}{2},\frac{1}{2},0\right)$ ...(ii)
Similarly for point $‘C’$
General point of line $\overrightarrow{r}=v\left(\hat{i}+\hat{j}+\hat{k}\right),\mathrm{ }C(v, v, v)$
Satisfy the plane $v+v+v=1$
$v=\frac{1}{3}$
Point $C\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ ....(iii)
$\Rightarrow \overrightarrow{AB}=-\frac{1}{2}\widehat{i}+\frac{1}{2}\widehat{j}$ and $\overrightarrow{AC}=-\frac{2}{3}\widehat{i}+\frac{1}{3}\widehat{j}+\frac{1}{3}\widehat{k}$
Now area of triangle $\Delta ABC=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|$
$\Delta =\left|\frac{1}{2}\left|\begin{array}{ccc}i& j& k\\ -\frac{1}{2}& \frac{1}{2}& 0\\ -\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right|\right|=\left|\frac{1}{2}\left(\frac{\hat{i}}{6}+\frac{\hat{j}}{6}+\frac{\hat{k}}{6}\right)\right|=\frac{\sqrt{3}}{12}$

$\Rightarrow {\left(6∆\right)}^2=\frac{3}{4}=0.75$