Given three lines

$$r = \lambda \widehat i,\,\lambda \in R$$, $$r = \mu (\widehat i + \widehat j),\,\mu \in R$$ and $$r = v(\widehat i + \widehat j + \widehat k),\,v\, \in R$$

cuts the plane x + y + z = 1 at the points A, B and C, respectively. So, for point A, put ($$\lambda $$, 0, 0) in the plane, we get $$\lambda $$ + 0 + 0 = 1 $$ \Rightarrow $$ $$\lambda $$ = 1 $$ \Rightarrow $$ A $$ \equiv $$ (1, 0, 0).

Similarly, for point B, put ($$\mu $$, $$\mu $$, 0) in the plane, we get $$\mu $$ + $$\mu $$ + 0 = 1 $$ \Rightarrow $$ $$\mu $$ = $${1 \over 2}$$

$$ \Rightarrow $$ B $$ \equiv $$ $$\left( {{1 \over 2},{1 \over 2},0} \right)$$.

and for point C, p;ut (v, v, v) in the plane we get

v + v + v = 1 $$ \Rightarrow $$ v = $${{1 \over 3}}$$ $$ \Rightarrow $$ C $$ \equiv $$ $$\left( {{1 \over 3},{1 \over 3},{1 \over 3}} \right)$$

Now, are of $$\Delta $$ABC = $${{1 \over 2}|AB \times AC| = \Delta }$$

$$ \because $$ AB = $${ - {1 \over 2}\widehat i + {1 \over 2}\widehat j}$$,

and AC = $${ - {2 \over 3}\widehat i + {1 \over 3}\widehat j + {1 \over 3}\widehat k}$$

$$ \therefore $$ AB $$ \times $$ AC = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1/2} & {1/2} & 0 \cr { - 2/3} & {1/3} & {1/3} \cr } } \right|$$

= $$\widehat i\left( {{1 \over 6}} \right) - \widehat j\left( { - {1 \over 6}} \right) + \widehat k\left( { - {1 \over 6} + {2 \over 6}} \right)$$

$$ = {1 \over 6}(\widehat i + \widehat j + \widehat k)$$

$$ \Rightarrow |AB \times AC| = {1 \over 6}\sqrt 3 = {1 \over {2\sqrt 3 }}$$

$$ \Rightarrow \Delta = {1 \over {4\sqrt 3 }}$$

$$ \Rightarrow {(6\Delta )^2} = 36{1 \over {16 \times 3}} = {3 \over 4} = 0.75$$