Equations of the plane containing the straight line x/2=y/3=z/4 and p…

The equation of a plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ is

$ ax + by + cz = 0 $ ..............(i)

where $2a + 3b + 4c = 0 $ ............(ii)

Let the direction ratios of the normal 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}$ be the proportional to $\alpha , \beta , \gamma $. Then, 

$ 3 \alpha + 4\beta + 2 \gamma = 0 $

and, $ 4\alpha + 2 \beta + 3 \gamma = 0 $

$∴ \frac{\alpha }{8}=\frac{\beta}{-1}=\frac{\gamma }{-10}$

It is given that the plane (i) is perpendicular to the plane the direction, ratios of normal to which are proportional to 8, -1, -10.

$∴ 8a - b - 10c = 0 $ .........(iii)

From (i) and (iii), we have

$\frac{a}{-30+4}=\frac{b}{32+20}=\frac{c}{-2-24}$

$⇒\frac{a}{-26}=\frac{b}{52}=\frac{c}{-26}$

$⇒ \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

Substituting the values of a, b, c in (ii), we obtain x - 2y + z = 0 as the required plane.