Equation of the plane containing the straight line `x/2=y/3=z/4` and perpendicular to the plane containing the straight lines `x/2=y/4=z/2` and `x/4=y/2=z/3` is

Explanation

Let the direction ratio of normal of the required plane is (a,b,c). So, the equation with the line contained in the plane will be:

`2a+3b+4c=0->Eq(1)`

Now, let direction ratio of normal of the plane perpendicular to the first plane is `(a_1,b_1,b_3)`.Then, lines with this plane will have equation:

`3a_1+4b_1+2c_1 = 0` and

`4a_1+2b_1+3c_1 = 0`

So, from above equations, we can find out the the direction ratios.

`a_1/(4**3-2**2) = -b_1/(3**3-2**4) = c_1/(3**2-4**4)`

`a_1/8 = -b_1/1 = c_1/-10`

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