Question

If a plane passes through a fixed point $(2,3,4)$ and meets the axes of reference in $A$, $B$ and $C$, the point of intersection of the planes through $A$, $B$, $C$ parallel to the coordinate planes can be

• A. $(6,9,12)$
• B. $(4,12,16)$
• C. $(1,1,-1)$
• D. $(2,3,-4)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(6,9,12)$
B $(1,1,-1)$
C $(4,12,16)$
D $(2,3,-4)$

Let us say a plane P $ax+by+cz=k$ passes through $(2,3,4)$ so $2a+3b+4c=k-\left(1\right)$

$A(\frac{k}{a},0,0),B(0,\frac{k}{b},0),C(0,0,\frac{k}{c})$

Points of intersection will be $⟨\frac{k}{a},\frac{k}{b},\frac{k}{c}⟩$

Let $\frac{k}{a}=x\frac{k}{b}=y\frac{k}{c}=z$ so in $\left(1\right)$

$\frac{2k}{x}+\frac{3k}{y}+\frac{4k}{z}=k$

$\frac{2}{x}+\frac{3}{y}+\frac{4}{z}=1-\left(1\right)$

$\left(a\right)$ if $(x,y,z)=(6,9,12)$

$\frac{2}{6}+\frac{3}{9}+\frac{4}{12}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ Hence true.

$\left(b\right)$ $⟨4,12,16⟩$

$\frac{2}{4}+\frac{3}{12}+\frac{4}{16}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1$ Hence correct

$\left(c\right)$ $⟨1,1,-1⟩$

$\frac{2}{1}+\frac{3}{1}+\frac{4}{-1}=1$ Hence this is also correct.

$\left(d\right)$ $⟨2,3,-4⟩$

$\frac{2}{2}+\frac{3}{3}+\frac{4}{-4}=2-1=1$ This is also correct.