Question

The distance of a point $(1,-2,3)$ from the plane $x-y+z=5$ and parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is

• A. 7
• B. 1
• C. 3
• D. 13

Answer 1

The correct option is B 1

Let the given point be $P(1,-2,3)$

And,

let $Q(x_1,y_1,z_1)$ be a point on the given plane

So, that $PQ$ is parallel to given line

then,

$(x_1,y_1,z_1)$ will lies on the given plane

So,

$x_1-y_1+z_1=5$ ----- $\left(1\right)$

Direction ratios of the line $PQ$ are $x_1-1,y_1+2,z_1-3$

Since,

Line $PQ$ is parallel to the line

$\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$

So,

their direction ratios will be proportion

i.e.,

$\frac{x_1-1}{2}=\frac{y_1+2}{3}=\frac{z_1-3}{-6}=k$ (say)

$x_1=2k+1,y_1=3k-2,z_1=-6k+3$

Putting the values of $x_1,y_1,z_1$ in $\left(1\right)$.

$\begin{array}{c}2k+1-\left(3k-2\right)+\left(-6k+3\right)=5\\ \\ 2k+1-3k+2-6k+3=5\\ \\ -7k+1=0\\ \\ k=\frac{1}{7}\end{array}$

$x_1=2k+1=\frac{9}{7}$

$y_1=3k-2=\frac{-11}{7}$

$z_1=-6k+3=\frac{15}{7}$

Thus coordinates of point $Q$ are

$(\frac{9}{7},\frac{-11}{7},\frac{-11}{4})$

Required distance, $=PQ$

$=\sqrt{(\frac{9}{7}-1{)}^2+(\frac{-11}{7}+2{)}^2+(\frac{15}{7}-3{)}^2}=1$

Distance $=1sq.units$

Hence the correct answer is $B$.