Question

Equation of the plane passing through the intersection of the planes $x+y+z=6$ and $2x+3y+4z+5=0$ and the point $(1, 1, 1)$ is

• A. $20x+23y+26z-69=0$
• B. $31x+45y+49z+52=0$
• C. $8x+5y+2z-69=0$
• D. $4x+5y+6z-7=0$

Answer 1

Answer: (A)

$20x+23y+26z-69=0$

Answer 2

Equation of plane containing plane $x+y+z-6=0$ and $2x+3y+4z+5=0$ is $(x+y+z-6)+X(2x+3y+4z+5)=0$ . Since, it passes through (1, 1, 1).
$\therefore$ $-3+\lambda .14=0$
$\Rightarrow$ $\lambda =\frac{3}{14}$
$\therefore$ Required equation of plane is $x+y+z-6+\frac{3}{14}(2x+3y+4z+5)=0$
$\Rightarrow$ $20x+23y+26z-69=0$