Equation of three lines `vecr=lamdahati,vecr=mu(hati+hatj),vecr=gamma(hati+hatj+hatk)` and a plane `x+y+z=1` are given
then area of triangle formed by point of intersectioin of line and plane is `Delta `then `(6Delta)^(2) `equals

Answer

0.75

Explanation

Put `(lamda,0,0)` in `x+y+z=1" "implieslamda=1impliesP(1,0,0)`
Put `(mu,mu,0)implies2mu=1impliesQ((1)/(2),(1)/(2),0)`
Put `(gamma,gamma,gamma)impliesgamma=(1)/(3)impliesR((1)/(3),(1)/(3),(1)/(3))`
Area of triangle `PQR=(1)/(2)|vec(PQ)xxvec(PR)|=(1)/(2)|((hati-hatj)/(2))xx((2hati-hatj-hatk)/(2))|=(1)/(12)|hati+hatj+hatk|=(sqrt(3))/(12)implies(6Delta)^(2)=0.75`