Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the point on the given line be $Q$. The coordinates of $Q$ can be written as $(2\lambda +1,4\lambda +3,3\lambda +2)$.

The direction ratios of the line segment $PQ$ are $(2\lambda +1-3,4\lambda +3-8,3\lambda +2-2)=(2\lambda -2,4\lambda -5,3\lambda )$.

Since the distance is measured parallel to the plane $3x+2y-2z+15=0$, the line $PQ$ is parallel to the plane. Therefore, the normal to the plane is perpendicular to $PQ$.

The direction ratios of the normal to the plane are $(3,2,-2)$.

Using the condition of perpendicularity:

$3(2\lambda -2)+2(4\lambda -5)-2(3\lambda )=0$

$6\lambda -6+8\lambda -10-6\lambda =0$

$8\lambda -16=0\Rightarrow \lambda =2$

Substituting $\lambda =2$, the coordinates of $Q$ are $(2(2)+1,4(2)+3,3(2)+2)=(5,11,8)$.

The distance $PQ$ is given by the distance formula:

$PQ=\sqrt{(5-3{)}^2+(11-8{)}^2+(8-2{)}^2}$

$PQ=\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$ units.

Answer: 7 units