Equation of plane passing through $(1, - 3, - 2)$ and perpendicular to planes $x + 2 y + 2 z = 5$ and $3 x + 3 y + 2 z = 8$ is

2 x + 4 y + 3 z + 8 = 0

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2 x - 4 y + 3 z - 8 = 0

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2 x - 4 y + 3 z + 8 = 0

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2 x - 4 y - 3 z - 8 = 0

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Solution

The correct option is B $2 x - 4 y + 3 z - 8 = 0$
Given planes: $P_{1} : x + 2 y + 2 z = 5$ and $P_{2} : 3 x + 3 y + 2 z = 8$
normal to $P_{1} :_{1} =^i + 2^j + 2^k$
normal to $P_{2} :_{2} = 3^i + 3^j + 2^k$
$Normal vector of required plane = {\to n}_{1} \times {\to n}_{2}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 2 & 2 3 & 3 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 2^i + 4^j - 3^k$

$So, Equation of plane - 2 x + 4 y - 3 z = d passing through (1, - 3, - 2) d = - 8 - 2 x + 4 y - 3 z = - 8 2 x - 4 y + 3 z - 8 = 0$

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Similar questions

(1, - 1, 2)

x + 2 y + 2 z = 5

3 x + 3 y + 2 z = 8

(- 1, 3, 2)

x + 2 y + 2 z = 5, 3 x + 3 y + 2 z = 8

2 x + 3 y - z + 1 = 0

x + y - 2 z + 3 = 0

3 x - y - 2 z - 4 = 0

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