JEE Advanced · MathematicsMedium

$A plane passes through the point (1, 1, 1) and is perpendicular to the planes x + 2 y + 3 z = 0 and $3x + 2y + z = 0$. Find the equation of the plane.$

Show correct answer & step-by-step solution

Correct answer: A — $x - 2 y + z = 0$

Solution

$The normals are n_{1} = (1, 2, 3) and n_{2} = (3, 2, 4) .$
$The normal to the required plane is n = n_{1} \times n_{2} = (2 - 6, 9 - 1, 2 - 6) = (- 4, 8, - 4), which is proportional to (1, - 2, 1) .$
$The equation is $1(x-1) - 2(y-1) + 1(z-1) = 0$.$
$Simplifying gives x - 2 y + z = 0 .$
$Hence the answer is (A).$

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