If the points (1, 2, 3) and (2, -1, 0) lie on the opposite sides of the plane $2 x + 3 y - 2 z = k$ , then

k < 1

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k > 2

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k < 1

k > 2

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1 < k < 2

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Solution

The correct option is D $1 < k < 2$
The points (1, 2, 3) and (2, -1, 0) lie on the opposite sides of the plane 2x + 3y - 2z - k = 0
So, $(1 \times 2 + 2 \times 3 + 3 \times (- 2) - k) (2 \times 2 + (- 1) \times 3 + 0 \times (- 2) - k)$
$⟹ (2 + 6 - 6 - k) (4 - 3 - k) < 0$
$\Rightarrow (k - 1) (k - 2) < 0$ ...... (i)
$∴ 1 < k < 2$
Hence, option D is correct.

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If the points (1, 2, 3) and (2, -1, 0) lie on the opposite sides of the plane 2x+3y−2z=k, then

The correct option is D 1<k<2The points (1, 2, 3) and (2, -1, 0) lie on the opposite sides of the plane 2x + 3y - 2z - k = 0So, (1×2+2×3+3×(−2)−k)(2×2+(−1)×3+0×(−2)−k)⟹(2+6−6−k)(4−3−k)<0⇒(k−1)(k−2)<0 ...... (i)∴1<k<2Hence, option D is correct.

If the points (1, 2, 3) and (2, -1, 0) lie on the opposite sides of the plane $2 x + 3 y - 2 z = k$ , then