If the equation of the plane through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2 x + 6 y + 6 z = 9$ is $p x + q y + r z = 9,$ then $p + q + r =$

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Solution

Let point : $A (2, 2, 1), B (9, 3, 6)$
$\Rightarrow - - \to A B = 7^i +^j + 5^k$
and plane $2 x + 6 y + 6 z = 9$ is perpendicular to the plane,
$\Rightarrow$ normal of plane is also perpendicular,
So, equation of plane is :
$∣ ∣ ∣ ∣ \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} a & b & c \end{matrix} ∣ ∣ ∣ ∣ = 0$
here, $(a, b, c) = (2, 6, 6)$ is the $D . r^{'} s$ of given plane.
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x - 2 & y - 2 & z - 1 7 & 1 & 5 2 & 6 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow 3 x + 4 y - 5 z = 9 \Rightarrow p = 3, q = 4, r = - 5$

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