⇒Line and Plane

Correct answer option is (B)

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}=\mathrm{k}$ (say)

Let $\mathrm{P}$ be the any point on the above line.

$\therefore \mathrm{P}\equiv (2\mathrm{k}+1,3\mathrm{k}-1,4\mathrm{k}+2)$

The point $\mathrm{P}$ lies on the plane

$\begin{array}{ll}\therefore & 2k+1+2(3k-1)+3(4k+2)=15\\ \therefore & 2k+1+6k-2+12k+6=15\\ \therefore & 20k=10\\ \therefore & k=\frac{1}{2}\\ \therefore & P\equiv (2,\frac{1}{2},4)\end{array}$

Distance of $\mathrm{P}$ from origin is

$\sqrt{2^2+{\left(\frac{1}{2}\right)}^2+4^2}=\sqrt{\frac{81}{4}}=\frac{9}{2}$

Correct answer option is (B)

Length of perpendicular from point $O(0,0,0)$ to plane $x-3y+4z-6=0$ is given by

$\begin{array}{rl}d& =\left|\frac{0(1)+0(-3)+0(4)-6}{\sqrt{(1{)}^2+(-3{)}^2+(4{)}^2}}\right|\\ & =\left|\frac{-6}{\sqrt{1+9+16}}\right|=\frac{6}{\sqrt{26}}\end{array}$