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# Answer please! - The equation of the plane containing the straight line and the perpendicular to the plane containin - Three Dimensional Geometry - JEE Main

The equation of the plane containing the straight line $x/2=y/3=z/4$ and the perpendicular to the plane containing the straight lines $x/3=y/4=z/2$ and $x/4=y/2=z/3$ is:

• Option 1)

$5x+2y-4z=0$

• Option 2)

$x+2y-2z=0$

• Option 3)

$x-2y+z=0$

• Option 4)

$3x+2y-3z=0$

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Equation of any plane passing through the line of intersection of two planes (Cartesian form ) -

The equation of any plane passing through the line of intersection of two planes

$ax+by+cz+d= 0$ and

$a_{1}x+b_{1}y+c_{1}z+d_{1}= 0$ is given by

$\left ( ax+by+cz+d \right )+\lambda \left ( a_{1}x+b_{1}y+c_{1}z+d _{1}\right )= 0$

-

From the concept we have learned vector along the normal to the plane containing the lines

$\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is $(8\hat{i} - \hat{j} - 10\hat{k})$

vector perpendicular to the vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $8\hat{i} -\hat{j} -10\hat{k}$  is  $26\hat{i} -52\hat{j} +26\hat{k}$

So, the required plane is

$26x -52y +26z = 0$ or  $x -2y +z = 0$

Option 1)

$5x+2y-4z=0$

Option 2)

$x+2y-2z=0$

Option 3)

$x-2y+z=0$

Option 4)

$3x+2y-3z=0$

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