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Let the equation of the plane passing through the line $x - 2 y - z - 5 = 0 = x + y + 3 z - 5$ and parallel to the line $x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2$ be $a x + b y + c z = 65 .$ Then the distance of the point $(a, b, c)$ from the plane $2 x + 2 y - z + 16 = 0$ is ……

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Detailed Solution

$x - 2 y - z - 5 + λ (x + y + 3 z - 5) = 0 \begin{array}{l} DR' S_{1} : (1 + λ, λ - 2, 3 λ - 1) \\ DR' S_{2} : (- 5, 3, 1) \end{array} WKT : - 5 (1 + λ) + 3 (λ + 2) + 1 (3 λ - 1) = 0 \Rightarrow λ = 12 ∴ P \equiv 13 x + 10 y + 35 z = 65$
$(a, b, c) = (13, 10, 35) ∴ d \equiv \frac{|26 + 20 - 35 + 16|}{3} = 9$

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Let the equation of the plane passing through the line x−2y−z−5=0=x+y+3z−5 and parallel to the line x+y+2z-7=0=2x+3y+z-2 be ax + by + cz = 65. Then the distance of the point (a, b, c) from the plane 2x + 2y – z + 16 = 0 is ……