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Question

Find the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = O.


Solution

Any plane through the given point (- 1, 3, 2) is
a(x - (-1)) + b (y - 3) + c(z - 2) = 0 or a(x + 1) + b (y - 3) + c(z -2) = 0 ...(i)
This plane (i) is at right angle to given plane x + 2y + 3z = 5
     a.1+ b.2 + c.3 = 0 x + 2b + 3c = 5 ...(ii)
Also, plane (i) is at right angle to given plane 3x + 3y + z =0
a.3 + b.3 + c.1= 3a + 3b + c =0         . . . (iii)
On solving Eqs. (ii) and we get
a2×13×3=b3×31×1=c1×32×3a7=b8=c3=λ    a=7λ, b=8λ,c=3λ
Putting the values of a, b, c in Eq. (i), we obtain
7λ(x+1)+8λ(y3)3λ(z2)=0
    (-7x -7) + (8y -24) - 3z + 6 =0
    - 7x + 8y - 3z -25 =0   7x -8y + 3z + 25 = 0
Hence, the equation of the required plane is 7x - 8y + 3z + 25 =0.

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