If the distance of the point (4, 6, 8) from the plane $\vec r. (6\hat i-12\hat j+4\hat k)= a$ is 1, then $a$ is:

-2 or -30

Point (4, 6, 8) and $\vec r. (6\hat i-12\hat j+4\hat k)= a$ 

$⇒(x\hat i+y\hat j+2\hat k)(6\hat i-12\hat j+4\hat k)= a$ 

$⇒6x-12y+4z-a=0$ equation of plane

distance of P from plane = 1 = $\frac{|6×4-12×6+4×8-a|}{\sqrt{6^2+(-12)^2+4^2}}$

∵ for plane $Ax+By+Cz+D=0$, Point $(x_0,y_0,z_0)$

distance given by → $\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$

$\frac{|24-72+32-a|}{\sqrt{36+144+16}}=1$

so $\frac{|16-a|}{14}=1$

so $|-16-a|=±14$

$-16-a=±14$

so $a=-16\mp 14$

$a=-2, -30$