Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The equation of any plane containing the given line is

$$\begin{array}{rlr}& (x+y+2z-3)+\lambda (2x+3y+4z-4)=0& \end{array}$$

$$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(2+4\lambda )z-(3+4\lambda )=0\dots \dots \left(i\right)$$

If the plane is parallel to $z$-axis whose direction cosines are $0,0,1$; then the normal to the plane will be perpendicular to $z$-axis

$$\begin{array}{llc}\therefore & (1+2\lambda )(0)+(1+3\lambda )(0)+(2+4\lambda )(1)=0\Rightarrow & \lambda =-\frac{1}{2}\end{array}$$

Put in eq. (1), the required plane is

$$\begin{array}{rlrl}& (x+y+2z-3)-\frac{1}{2}(2x+3y+4z-4)=0\Rightarrow y+2=0\dots (2)& & \end{array}$$

$$\therefore S.D.=distanceofanypointsay(0,0,0)onz-axisfromplane(2)=\frac{2}{\sqrt{(1{)}^2}}=2$$