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  4. If the equation of a plane P, passing through the intersection of the planes, x+4 y-z+7=0 and 3 x+y+5 z=8 is a x+b y+6 z=15 for some a, b ∈ R, then the distance of the point (3,2,-1) from the plane P is .

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Q. If the equation of a plane P, passing through the intersection of the planes, $x+4 y-z+7=0$ and $3 x+y+5 z=8$ is $a x+b y+6 z=15$ for some $a, b \in R,$ then the distance of the point (3,2,-1) from the plane $P$ is _______.

JEE MainJEE Main 2020Three Dimensional Geometry

Solution:

$D_{1} = \begin{vmatrix}-7&4&-1\\ 8&1&5\\ 15&b&6\end{vmatrix}= 0$
$\Rightarrow b = -3 $
$D = \begin{vmatrix}1&4&-1\\ 3&1&5\\ a&b&6\end{vmatrix} = 0 $
$\Rightarrow 21 a - 8b - 66 = 0 \dots \left(1\right) $
$P : 2x - 3y + 6z = 15$
so required distance $= \frac{21}{7} = 3$