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The plane denoted by $Π_{1} : 4 x + 7 y + 4 z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $Π_{2} : 5 x + 3 y + 10 z = 25 .$ If the plane in its new position be denoted by $Π,$ and the distance of this plane from the origin is $\sqrt{k}$ where $k \in N,$ then find k.

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answer is 212.

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Detailed Solution

Equation of the plane P is $4 x + 7 y + 4 z + 81 + λ (5 x + 3 y + 10 z - 25) = 0$

$(5 λ + 4) x + (3 λ + 7) y + (10 λ + 4) z + (81 - 25 λ) = 0$

Now, $4 (5 λ + 4) + 7 (3 λ + 7) + (10 λ + 4) 4 = 0$

$λ = - 1$

Equation of the plane $- x + 4 y - 6 z + 106 = 0$

$k = 212$

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