Question

Find the co-ordinates of a point lying on the line $\frac{x-2}{3}=\frac{y+3}{4}=\frac{z-1}{7}$ which is at a distance $10$ units from $(2,-3,1)$.

• A. $(32,37,71)$
• B. $(-28,-43,-69)$
• C. $(-32,-37,-71)$
• D. None of these

Answer 1

The correct option is D None of these

Given that the required point lies on the line $\frac{x-2}{3}=\frac{y+3}{4}=\frac{z-1}{7}$.

The required point is at a distance of $10$ units from $(2,-3,1)$

By option verification,

(A) Substituting the point $(32,37,71)$

$⟹\frac{32-2}{3}=\frac{37+3}{4}=\frac{71-1}{7}$

$⟹10=10=10$

Therefore, distance is $\sqrt{(32-2{)}^2+(37+3{)}^2+(71-1{)}^2}=86.02$ units

Hence, $(32,37,71)$ is not the required point.

(B) Substituting the point $(-28,-43,-69)$

$⟹\frac{-28-2}{3}=\frac{-43+3}{4}=\frac{-69-1}{7}$

$⟹-10=-10=-10$

Therefore, distance is $\sqrt{(-28-2{)}^2+(-43+3{)}^2+(-69-1{)}^2}=86.02$ units

Hence, $(-28,-43,-69)$ is not the required point.

(C) Substituting the point $(-32,-37,-71)$

$⟹\frac{-32-2}{3}=\frac{-37+3}{4}=\frac{-71-1}{7}$

$⟹11.3=11.3=10.28$

Therefore, distance is $\sqrt{(-32-2{)}^2+(-37+3{)}^2+(-71-1{)}^2}=86.57$ units

Hence, $(-32,-37,-71)$ is not the required point.