The equation of any plane through the line of intersection of the planes and is given by -

⇒

So, the equation of any plane through the line of intersection of the given planes is

⇒

⇒

∴ …(1)

Since this plane is perpendicular to the plane

…(2)

So, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).

Direction ratios of Normal of plane (1) = (a_1, b_1, c₁)

≡ [(1 - 2λ), (2 - λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a_2, b_2, c₂)

≡ (-5, -3,6)

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒ (1 - 2λ) × (-5) + (2 - λ) × (-3) + (3 + λ) × 6 = 0

⇒ -5 + 10λ - 6 + 3λ + 18 + 6λ = 0

⇒ 19λ + 7 = 0

∴ λ = -7/19

Putting the value of λ in (1), we get -

⇒

⇒

⇒

Hence, the equation of the required plane is