Question

The variable plane $(2\lambda +1)x+(3-\lambda )y+z=4$ always passes through the line

• A. $\frac{x}{0}=\frac{y}{0}=\frac{x+4}{1}$
• B. $\frac{x}{1}=\frac{y}{2}=\frac{z}{-3}$
• C. $\frac{x}{1}=\frac{y}{2}=\frac{z-4}{-7}$
• D. none of these

Answer 1

Answer: (C)

$\frac{x}{1}=\frac{y}{2}=\frac{z-4}{-7}$

Given equation of plane is $(2\lambda +1)x+(3-\lambda )y+z=4$

Here $\lambda$ is a variable, we have to eliminate it.

The given equation can be written as $\lambda (2x-y)+x+3y+z=4$

Now observe that $\lambda$ will be eliminated from the equation, if $2x-y=0$

$\Rightarrow 2x=y$

$\Rightarrow x=\frac{y}{2}$

The given equation becomes $x+3y+z=4$

$\Rightarrow 7x+z=4$

$\Rightarrow x=\frac{z-4}{-7}$

So, the given plane always passes through $\frac{x}{1}=\frac{y}{2}=\frac{z-4}{-7}$ irrespective of $\lambda$

Therefore, option $C$ is correct.