Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the three points be $A(2,1,-1)$, $B(1,3,2)$, and $C(1,2,1)$.

The plane passes through these three points. We can find two vectors lying in the plane:
$\overrightarrow{AB}=(1-2)\hat{i}+(3-1)\hat{j}+(2-(-1))\hat{k}=-\hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{AC}=(1-2)\hat{i}+(2-1)\hat{j}+(1-(-1))\hat{k}=-\hat{i}+\hat{j}+2\hat{k}$

The normal vector $\vec{n}$ to the plane is given by the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$:
$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 2& 3\\ -1& 1& 2\end{array}\right|$

$\vec{n}=\hat{i}(4-3)-\hat{j}(-2-(-3))+\hat{k}(-1-(-2))$
$\vec{n}=\hat{i}-\hat{j}+\hat{k}$

The equation of the plane passing through $C(1,2,1)$ with normal vector $(1,-1,1)$ is:
$1(x-1)-1(y-2)+1(z-1)=0$
$x-1-y+2+z-1=0$
$x-y+z=0$

To find the intercepts on the coordinate axes, we substitute $y=0,z=0$ for the x-intercept, $x=0,z=0$ for the y-intercept, and $x=0,y=0$ for the z-intercept.
For the x-intercept $p$: $p-0+0=0\Rightarrow p=0$
For the y-intercept $q$: $0-q+0=0\Rightarrow q=0$
For the z-intercept $r$: $0-0+r=0\Rightarrow r=0$

Therefore, the sum of the intercepts is:
$\mathrm{p}+\mathrm{q}+\mathrm{r}=0+0+0=0$

Answer: $0$