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

Answer:

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)#