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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Question 10 Find the vector equation of the plane passing through the intersection of the planes π β . (π Μ + π Μ + π Μ) = 6 and π β . (2π Μ + 3π Μ + 4π Μ) = β 5, and the point (1, 1, 1).The vector equation of a plane passing through the intersection of planes π β. (π1) β = d1 and π β. (π2) β = d2 and also through the point (x1, y1, z1) is π β.((ππ) β + π(ππ) β) = d1 + πd2 Given, the plane passes through π β.(π Μ + π Μ + π Μ) = 6 Comparing with π β.(π1) β = d1, (ππ) β = π Μ + π Μ + π Μ & d1 = 6 π β.(2π Μ + 3π Μ + 4π Μ) = β5 βπ β.(2π Μ + 3π Μ + 4π Μ) = 5 π β .(β 2π Μ β 3π Μ β 4π Μ) = 5 Comparing with π β.(π2) β = d2 (ππ) β = β 2π Μ β 3π Μ β 4π Μ & d2 = 5 Equation of plane is π β. [(π Μ+π Μ+π Μ )+"π" (β2π Μβ3π Μβ4π Μ)] = 6 + π5 π β. [(π Μ" " +π Μ" " +π Μ )β"π" (ππ Μ+ππ Μ+ππ Μ)] = 6 + 5π Now to find π , put π β = xπ Μ + yπ Μ + zπ Μ (xπ Μ + yπ Μ + zπ Μ). [(π Μ+π Μ+π Μ )β"π" (2π Μ+3π Μ+4π Μ)] = 5π + 6 (xπ Μ + yπ Μ + zπ Μ).(π Μ+π Μ+π Μ ) β π (xπ Μ + yπ Μ + zπ Μ).(2π Μ+3π Μ+4π Μ) = 5π + 6 (x Γ 1) + (y Γ 1) + (z Γ 1) β π[(π₯Γ2)+(π¦Γ3)+(π§Γ4)] = 5π + 6 x + y + z β π[2π₯+3π¦+4π§] = 5π + 6 x + y + z β 2ππ₯ β 3πy β 4πz = 5π + 6 (1 β 2π)x + (1 β 3π)y + (1 β 4π) z = 5π + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 β 2π)x + (1 β 3π)y + (1 β 4π) z = 5π + 6 (1 β2π) Γ 1 + (1 β 3π) Γ 1 + (1 β 4π) Γ 1 = 5π + 6 1 β2π + 1 β 3π + 1 β 4π= 5π + 6 3 β 9π = 5π + 6 β14π = 3 β΄ π = (βπ)/ππ Putting value of π in (1), π β. [(π Μ" " +" " π Μ" " +" " π Μ )β(( β3)/14)(2π Μ+3π Μ+"4" π Μ)]= 6 + 5 Γ ( β3)/14 π β. [(π Μ+π Μ+" " π Μ )+3/14(2π Μ+3π Μ+"4" π Μ)]= 6 β 15/14 π β. [π Μ+π Μ" " +π Μ+ 6/14 π Μ+9/14 π Μ+12/14 π Μ ]= 69/14 π β. [(1+6/14) π Μ +(1+9/14) π Μ+(1+12/14) π Μ ]= 69/14 π β. [20/14 π Μ + 23/14 π Μ + 26/14 π Μ ]= 69/14 π β. [1/14(20π Μ+23π Μ+26π Μ)]= 69/14 1/14 π β. (20π Μ + 23π Μ + 26π Μ) = 69/14 π β. (20π Μ + 23π Μ + 26π Μ) = 69 Therefore, the vector equation of the required plane is π β.(πππ Μ + πππ Μ + πππ Μ) = ππ