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=?

1 Answer
Mar 22, 2018

See below.

Explanation:

If the planes

P -> x+ay-3z+3=0
Q->x+2y-bz+c=0

are parallel then their normal vectors are aligned or

(1,a,-3) = lambda(1,2,-b)

now the Q intercepts are

x-> -c, y->-c/2,z-> c/b

then

{(c(1/b-3/2)=14),(a = 2 lambda),(3 = b lambda):}

which is equivalent to

{(c(1/b-3/2)=14),(a/3 = 2/b):}

Solving for (a,b) we obtain

a = (3 (28 + 3 c))/c
b = (2 c)/(28 + 3 c)

and finally

a+b+c = 9 + 84/c + c + (2 c)/(28 + 3 c)