$A$ symmetrical form of the line of intersection of the planes $x = ay + b$ and $z = cy + d$ is

The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is:

The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $P: ax-y-z=0$,$(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta-\gamma$ is equal to $........$

Let the plane $P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$. If the intercept of $P$ on the $y$-axis is $1$,then the distance between $P$ and $L$ is:

The angle between the plane $x + y + z = 5$ and the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^{\circ}$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in its new position is: