QJEEAdv19Ma7cb6: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma7cb6 Solution: Three lines are given by , and Let the lines cut the plane x +…

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 = $$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1/2& 1/2& 0\\ -2/3& 1/3& 1/3\end{array}\right|$$

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

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

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

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

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