What is the distance between the line $r = (2i - 2j + 3k) + \lambda (i - j + 4k)$ and the plane $r \cdot (i + 5j + k) = 5$?

The distance of the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ from the point $(3, 4, 5)$ is given by

For non-coplanar vectors $a, b$ and $c$,if the point of intersection of the line $r=a+t(b-c)$ and the plane $r=b+c+x(a-b)+y(c+a)$ is $l a+m b+n c$,then $3 l+4 m+2 n=$

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then $(\alpha, \beta)=$

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$,be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$,is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$,then $\alpha + \beta + \gamma$ is equal to:

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $............$.