(a) Let $\mathrm{N}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ be the foot of the perpendicualr from the origin to the plane
$$
2 x+3 y+4 z-12=0
$$
$\therefore \quad$ Direction ratios of the normal are 2, 3, 4. Also the direction ratios of $\mathrm{ON}$ are $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$
$\Rightarrow \quad \frac{x_1}{2}=\frac{y_1}{3}=\frac{z_1}{4}=k$
$\therefore \quad x_1=2 k, y_1=3 k, z=4 k$
The point $\left(x_1, y_1, z_1\right)$ lies on the plane $\therefore 2(2 \mathrm{k})+3(3 \mathrm{k})+4(4 \mathrm{k})-12=0$
$$
(4+9+16) \mathrm{k}=12, \quad \therefore \mathrm{k}=\frac{12}{29}
$$
Putting value of $k$ in (i)
$$
\begin{aligned}
\mathrm{x}_1 &=2 \cdot \frac{12}{29}=\frac{24}{29}, \quad \mathrm{y}_1=3 \mathrm{k}=3 \cdot \frac{12}{29}=\frac{36}{29} \\
\therefore \quad \mathrm{z}_1 &=4 \mathrm{k}=4 \cdot \frac{12}{29}=\frac{48}{29}
\end{aligned}
$$
Hence, the foot of the normal from the origin to the given plane is $\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)$.
(b) $3 y+4 z-6=0$
Direction ratios of the normal to the plane are:
$$
\mathrm{a}=0, \mathrm{~b}=3, \mathrm{c}=4
$$
$\therefore \quad$ Equation of the line through $(0,0,0)$ are
$$
\begin{aligned}
&\quad \frac{x-0}{0}=\frac{y-0}{3}=\frac{z-0}{4} \Rightarrow \frac{x}{0}=\frac{y}{3}=\frac{z}{4}=\lambda \text { (let) } \\
&\therefore \quad \mathrm{x}=0, \mathrm{y}=3 \lambda, \mathrm{z}=4 \lambda \\
&\text { Let foot of the perpendicular from the origin be }(0,3 \lambda, 4 \lambda \text { ) } \\
&\text { which lies on the plane } \\
&\quad 3 \mathrm{y}+4 \mathrm{z}-6=0 \Rightarrow 3(3 \lambda)+4(4 \lambda)-6=0 \\
&\Rightarrow \quad \lambda=\frac{6}{25} \therefore \mathrm{x}=0, \mathrm{y}=\frac{18}{25}, \mathrm{z}=\frac{24}{25}
\end{aligned}
$$
Hence the foot of the perpendicular is $\left(0, \frac{18}{25}, \frac{24}{25}\right)$
(c) Direction ratios of the normal to the plane $x+y+z=1 a=1, b=1, c=1$
Now equation of the line through $(0,0,0)$ are:
$\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1} \Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=\lambda$ (let)
$\therefore \quad x=\lambda, y=\lambda, z=\lambda$
Let foot of the perpendicular from the origin be
$(\lambda, \lambda, \lambda$,$) which lies on the plane x+y+z=$
$\lambda+\lambda+\lambda=1 \Rightarrow \lambda=\frac{1}{3} \therefore x=\frac{1}{3}, \mathrm{y}=\frac{1}{3}, \mathrm{z}=\frac{1}{3}$
(d) $5 y+8=0$
Direction ratios of the normal to the plane are :
$$
\mathrm{a}=0, \mathrm{~b}=5, \mathrm{c}=0
$$
Now equation of the line through $(0,0,0)$ are:
$$
\begin{aligned}
& \frac{x-0}{0}=\frac{y-0}{5}=\frac{z-0}{0} \Rightarrow \frac{x}{0}=\frac{y}{5}=\frac{z}{0}=\lambda \text { (let) } \\
\therefore \quad & x=0, y=5 \lambda, \mathrm{z}=0 . \\
& \text { Let foot of the perpendicular be }(0,5 \lambda, 0) \text { which lies } \\
\text { on: } 5 \mathrm{y}+8=0 \\
& \Rightarrow 5(5 \lambda)+8=0 \Rightarrow \lambda=\frac{-8}{25} \\
\therefore \quad & \mathrm{x}=0, \mathrm{y}=5 \times\left(\frac{-8}{25}\right)=\frac{-8}{5}, \mathrm{z}=0
\end{aligned}
$$
Hence the foot of the perpendicular is $\left(0, \frac{-8}{5}, 0\right)$