A plane II makes intercepts 3 and 4 respectively on x and z axes. If…

Let the equation of the plane be

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ .................(i)

It makes intercepts 3 and 4 respectively on x and z-axis.

$∴ a = 3 $ and $ c = 4 $

It is given that the plane (i) is parallel to y -axis

i.e $\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$

$∴ \frac{1}{a}×0 +\frac{1}{b}×1+\frac{1}{c}×0 = 0 ⇒ \frac{1}{b}= 0 $

Hence, the equation of the plane is

$\frac{x}{3}+\frac{z}{4}= 1 $ or , $ 4x + 3z = 12 $