Answer
An equation of the plane is
$ax + by + \frac{1}{4}\left( {3a + b} \right)z = d$
Work Step by Step
By definition, the angle between two planes is the angle between their normal vectors.
The normal vector of the plane $3x + y - 4z = 2$ is ${{\bf{n}}_1} = \left( {3,1, - 4} \right)$.
Let ${{\bf{n}}_2} = \left( {a,b,c} \right)$ denote the normal vector of the plane required. Since ${{\bf{n}}_2}$ makes an angle $\frac{\pi }{2}$ with ${{\bf{n}}_1}$, their dot product vanishes, that is ${{\bf{n}}_1}\cdot{{\bf{n}}_2} = 0$. So,
$\left( {3,1, - 4} \right)\cdot\left( {a,b,c} \right) = 0$
$3a + b - 4c = 0$
$c = \frac{1}{4}\left( {3a + b} \right)$
So, ${{\bf{n}}_2} = \left( {a,b,\frac{1}{4}\left( {3a + b} \right)} \right)$.
The equation of the plane required is
${{\bf{n}}_2}\cdot\left( {x,y,z} \right) = d$
$\left( {a,b,\frac{1}{4}\left( {3a + b} \right)} \right)\cdot\left( {x,y,z} \right) = d$
$ax + by + \frac{1}{4}\left( {3a + b} \right)z = d$
Note that there are infinitely many such planes that make an angle of $\frac{\pi }{2}$ with the plane $3x+y-4z=2$, corresponding to the values of $a$, $b$ and $d$.
