Question

If the equation of the plane passing through $(4,0,1)$ and parallel to the plane $4x+3y-12z+6=0$ is $ax+by+cz=d,(a>0)$, then

• A. $ab=cd$
• B. $4ab-cd=0$
• C. $cd+4ab=0$
• D. $a+b+c+d=0$

Answer 1

Answer: (C)

$cd+4ab=0$

The equation of the plane parallel to the given plane will be $4x+3y-12z=d$.
Since, it passes through $(4,0,1)$, thus $d=4$
Now, the equation of plane becomes
$\left(4k\right)x+\left(3k\right)y-\left(12k\right)z=4k$, where $k\in R$
$\therefore a=4k,b=3k,c=-12k,d=4k$
$ab\ne cd$
$4ab-cd=0,$ for $k=0,$ which is not possible.
$cd+4ab=0$, for any value of $k$
$a+b+c+d=0$, for $k=0$, which is not possible.
Hence, option C is correct.