Question

Find the equation of the plane containing the line of intersetion of the plane $x+y+z-6=0$ and $2x+3y+4z-5=0$ and passing through $(1,1,1)$.

Answer 1

$P_1:x+y+z-6=0$

$P_2:2x+3y+z-5=0$

$\therefore$ plane containing line of intersection of $P_1$ and $P_2$ will be $P_1+\lambda P_2$

$\therefore$ required plane, $x+y+z-6+\lambda (2x+3y+4z-5)=0$

$(2\lambda +1)x+(3\lambda +1)y+(4\lambda +1)z-5\lambda -6=0$

it passes through $(1,1,1)$

$\therefore 2\lambda +1+3\lambda +1+4\lambda +1-5\lambda -6=0$

$4\lambda =3\Rightarrow \lambda =\frac{3}{4}$

Ans: $\frac{5}{2}x+\frac{13}{4}y+4z-\frac{39}{4}=0$