Question

The equation of the plane through the line of intersection of the planes $2x+y-z-4=0$ and $3x+5z-4=0$ which cuts off equal intercepts from the $x-$axis and the $y$-axis is

• A. $3x+3y-8z+8=0$
• B. $3x+3y-8z-8=0$
• C. $3x-3y-8z-8=0$
• D. $x+y-8z-8=0$

Answer 1

The correct option is A $3x+3y-8z+8=0$

The equation of any plane passing through the lines of intersection of the given planes

$2x+y-z-4=0$ and $3x+5z-4=0$ is

$(2x+y-z-4)+k(3x+5z-4)=0-\left(1\right)$

For the intercept on $x-axis$, on putting $y=0$ and $z=0$, we get

$(2x-4)+k(3x-4)=0$

$⟹x=\frac{4k+4}{3k+2}$

For the intercept on $y-axis$, on putting $x=0,z=0$, we get

$y(-4)+k(-4)=0$

$⟹y=4k+4$

Since the intercepts on x-axis and y-axis are equal.

$\therefore \frac{4k+4}{3k+2}=4k+4$

$⟹4k+4=12k^2+12k+8k+8$

$⟹12k^2+16k+4=0$

$⟹3k^2+4k+4=0$

$⟹k=\frac{1}{3}$

Therefore the requird plane is

$(2x+y-z-4)+\frac{1}{3}(3x+5z-4)=0$

$⟹3x+3y-8z+8=0$