# If P : x+ay-3z+3=0 and Q : x+2y-bz+c=0 are two parallel planes such that sum of intercepts made by Q on the axes is 14 then value of (a + b + c) will be=?

Mar 22, 2018

See below.

#### Explanation:

If the planes

$P \to x + a y - 3 z + 3 = 0$
$Q \to x + 2 y - b z + c = 0$

are parallel then their normal vectors are aligned or

$\left(1 , a , - 3\right) = \lambda \left(1 , 2 , - b\right)$

now the $Q$ intercepts are

$x \to - c , y \to - \frac{c}{2} , z \to \frac{c}{b}$

then

$\left\{\begin{matrix}c \left(\frac{1}{b} - \frac{3}{2}\right) = 14 \\ a = 2 \lambda \\ 3 = b \lambda\end{matrix}\right.$

which is equivalent to

$\left\{\begin{matrix}c \left(\frac{1}{b} - \frac{3}{2}\right) = 14 \\ \frac{a}{3} = \frac{2}{b}\end{matrix}\right.$

Solving for $\left(a , b\right)$ we obtain

$a = \frac{3 \left(28 + 3 c\right)}{c}$
$b = \frac{2 c}{28 + 3 c}$

and finally

$a + b + c = 9 + \frac{84}{c} + c + \frac{2 c}{28 + 3 c}$