Explanation

(b)
Plane passing through the line of intersection if planes $4y + 6\mathcal{z} = 5$ and $2x + 3y + 5\mathcal{z} = 5$ is $\left( 4y + 6\mathcal{z} - 5 \right) + \lambda\left( 2x + 3y + 5\mathcal{z} - 5 \right) = 0$, or
$2\lambda x + (3\lambda + 4)y + (5\lambda + 6)\mathcal{z} - 5\lambda - 5 = 0$
Clearly, for $\lambda = - 3$, we get the plane $6x + 5y + 9\mathcal{z} = 10$
Hence, the given three planes have common line of intersection