Question

Equation of the plane passing through the intersection of the planes $x+y+z=6$, $2x+3y+4z+5=0$ 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

The correct option is A

$20x+23y+26z–69=0$

Explanation for the correct option:

Given, equation of planes are

$x+y+z=6$

$2x+3y+4z+5=0$

Equation of plane passing through intersection of given plane is,

$L_1+\lambda L_2=0$

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

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

$\Rightarrow 1+1+1+-6+\lambda (2+3+4+5)=0$

$\Rightarrow$ $\lambda =\frac{3}{14}$
$\therefore$the required plane is,

$14(x+y+z-6)+3(2x+3y+4z+5)=0$

$\Rightarrow$ $\mathbf{20}\mathit{x}\mathbf{+}\mathbf{23}\mathit{y}\mathbf{+}\mathbf{26}\mathit{z}\mathbf{-}\mathbf{69}\mathbf{=}\mathbf{0}$

Hence, Option ‘A’ is Correct.