Question

Distance between two non-intersecting planes $P_{1}$ and $P_{2}$ is $5units$ , where $P_{1}$ is $2x-3y+6z+26=0$ and $P_{2}$ is $4x+by+cz+d=0.$ The point $A\left(- 3 , 0 , - 1\right)$ lies between the planes $P_{1}$ and $P_{2}$ , then the value of $3b+4c-5d$ is equal to

• A. $580$
• B. $120$
• C. $-18$
• D. $-120$

Answer 1

Answer: (B)

Since both the given planes are parallel.
So the two planes would be $P_{1}:4x-6y+12z+52=0$ and $P_{2}:4x-6y+12z+d=0$
Therefore, $\left|\frac{d - 52}{14}\right|=5\Rightarrow \left|d - 52\right|=70$
$\Rightarrow d=122,-18$
$\therefore P_{2}$ is $4x-6y+12z+122=0$ or $4x-6y+12z-18=0$
Since the point $\left(- 3,0 , - 1\right)$ is lying between $P_{1}$ and $P_{2}$
$\therefore $ On substituting the point in both the equations of the plane, both the expressions must be of opposite signs, which is satisfied by $4x-6y+12z-18=0$ .
Hence, $4x-6y+12z-18=0$ is the equation of the required plane