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Plane perpendicular to two given planes (normal = cross product of their normals)

This type of question, asked 5 times: 2021 Paper 1, 28 Jun 2022, 25 Jul 2022, 30 Jan 2023, 11 Apr 2023. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 1

    63
    The equation of the plane passing through the point (1,2,−3)(1,2, - 3) and perpendicular to the planes 3x+y−2z=53x+y- 2z= 5 and 2x−5y−z=72x- 5y-z= 7, is:

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  2. Question 2

    28 Jun 2022

    166
    Let the plane ax+by+cz=dax + by + cz = d pass through (2,3,−5)(2,3, - 5) and is perpendicular to the planes 2x+y−5z=102x + y - 5z = 10 and 3x+5y−7z=123x + 5y - 7z = 12. If a,b,c,da,b,c,d are integers d>0d > 0 and gcd⁡(∣a∣,∣b∣,∣c∣,d)=1\gcd(|a|,|b|,|c|,d)= 1, then the value of a+7b+c+20da + 7b + c + 20d is equal to

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  3. Question 3

    25 Jul 2022

    165
    A plane EE is perpendicular to the two planes 2x−2y+z=02x - 2y + z = 0 and x−y+2z=4x - y + 2z = 4, and passes through the point P(1,−1,1)P(1, - 1,1). If the distance of the plane EE from the point Q(a,a,2)Q(a,a,2) is 323\sqrt{2}, then (PQ)2(PQ)^{2} is equal to

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  4. Question 4

    30 Jan 2023

    85
    If the equation of the plane passing through the point (1,1,2)(1,1,2) and perpendicular to the line x−3y+x - 3y + 2z−1=0=4x−y+z2z - 1 = 0 = 4x - y + z is Ax+By+Cz=1Ax + By + Cz = 1, then 140(C−B+A)140(C - B + A) is equal to

  5. Question 5

    11 Apr 2023

    62
    If equation of the plane that contains the point (−2,3,5)( - 2,3,5) and is perpendicular to each of the planes 2x+4y+5z=82x + 4y + 5z = 8 and 3x−2y+3z=53x - 2y + 3z = 5 is αx+βy+γz+97=0\alpha x + \beta y + \gamma z + 97 = 0 then α+β+γ=\alpha + \beta + \gamma =

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