Question

The ratio in which the line joining $(2,-4,3)$ and $(-4,5,-6)$ is divided by the plane $3x+2y+z-4=0$ is

• A. $2:1$
• B. $4:3$
• C. $1:4$
• D. $2:3$

Answer 1

The correct option is C $1:4$

We have given points as shown in the figure

$Eq{u}^{n}oflineAB:\frac{x-2}{6}=\frac{y+4}{-9}=\frac{z-3}{9}$

Now

Points $P$ is $\left(6r+2,-9r-4,9r+3\right)$

$3\left(6r+2\right)+2\left(-9r-4\right)+9r+3=4$

$9r+1=4$

$\therefore r=\frac{1}{3}$

So,

$P\left(4,-7,6\right)$

Now,

$\frac{AP}{BP}=\sqrt{\frac{{\left(-2\right)}^2+{\left(3\right)}^2+{\left(-3\right)}^2}{{\left(8\right)}^2+{\left(12\right)}^2+{\left(12\right)}^2}}$

$=\sqrt{\frac{22}{352}}$

$=\sqrt{\frac{11}{176}}$

$=\sqrt{\frac{1}{16}}=\frac{1}{4}$

As $P$ divides externally in the ratio of $1:4$.

Hence, The option $C$ is the correct answer.