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

The equation of the line passing through the points $\mathrm{Q}(3,-4,-5)$ and $\mathrm{R}(2,-3,1)$ is given by:

$\frac{x-3}{2-3}=\frac{y-(-4)}{-3-(-4)}=\frac{z-(-5)}{1-(-5)}$

$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda$

Any point on this line can be represented as $(-\lambda +3,\lambda -4,6\lambda -5)$.

Since this point lies on the plane $2x+y+z=7$, we substitute these coordinates into the plane's equation:

$2(-\lambda +3)+(\lambda -4)+(6\lambda -5)=7$

$-2\lambda +6+\lambda -4+6\lambda -5=7$

$5\lambda -3=7$

$5\lambda =10\Rightarrow \lambda =2$

Substituting $\lambda =2$ back into the coordinates, we get the point of intersection:

$x=-2+3=1$

$y=2-4=-2$

$z=6(2)-5=7$

The point of intersection is $\mathrm{S}(1,-2,7)$.

The distance of the point $\mathrm{P}(3,4,4)$ from $\mathrm{S}(1,-2,7)$ is calculated using the distance formula:

$\mathrm{PS}=\sqrt{(3-1{)}^2+(4-(-2){)}^2+(4-7{)}^2}$

$\mathrm{PS}=\sqrt{2^2+6^2+(-3{)}^2}$

$\mathrm{PS}=\sqrt{4+36+9}=\sqrt{49}=7$ units

Answer: 7 units