Equation of the plane containing the line $x + 2 y + 3 z - 5 = 0 = 3 x + 2 y + z - 5$ which is parallel to the line $x - 1 = 2 - y = z - 3$ , is-

x + 2 y + z = 0

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x - z = 0

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2 y + 2 z = 1

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x + 4 y + 3 z = 0

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Solution

The correct option is B $x - z = 0$
Let plane is -

$(x + 2 y + 3 z - 5) + λ (3 x + 2 y + z - 5) = 0$

Plane is parallel to $\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{1}$

$\Rightarrow 1 (1 + 3 λ) - 1 (2 + 2 λ) + 1 (3 + λ) = 0$

$\Rightarrow λ = - 1$

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Similar questions

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\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

and perpendicular to the plane containing the straight lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3}

is :

Q. The equation of the plane containing the straight line

\frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{z}{5}

and perpendicular to the plane

x - y + z + 2 = 0

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Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

Q. The equation of the plane passing through the point

(- 1, 2, 0)

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Equation of the plane containing the line x+2y+3z−5=0=3x+2y+z−5 which is parallel to the line x−1=2−y=z−3, is-

The correct option is B x−z=0Let plane is - (x+2y+3z−5)+λ(3x+2y+z−5)=0 Plane is parallel to x−11=y−2−1=z−31 ⇒1(1+3λ)−1(2+2λ)+1(3+λ)=0 ⇒λ=−1

Equation of the plane containing the line $x + 2 y + 3 z - 5 = 0 = 3 x + 2 y + z - 5$ which is parallel to the line $x - 1 = 2 - y = z - 3$ , is-

The correct option is B $x - z = 0$
Let plane is -

$(x + 2 y + 3 z - 5) + λ (3 x + 2 y + z - 5) = 0$

Plane is parallel to $\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{1}$

$\Rightarrow 1 (1 + 3 λ) - 1 (2 + 2 λ) + 1 (3 + λ) = 0$

$\Rightarrow λ = - 1$