Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The plane passes through the point $P(0,4,3)$ and contains the line $\frac{x-1}{1}=\frac{y-3}{3}=\frac{z-2}{2}$.

The line passes through the point $Q(1,3,2)$ and has a direction vector $\vec{v}=\hat{i}+3\hat{j}+2\hat{k}$.

The vector joining the points $P$ and $Q$ lies on the plane:
$\overrightarrow{PQ}=(1-0)\hat{i}+(3-4)\hat{j}+(2-3)\hat{k}=\hat{i}-\hat{j}-\hat{k}$

The normal vector $\vec{n}$ to the plane is perpendicular to both $\overrightarrow{PQ}$ and $\vec{v}$:
$\vec{n}=\overrightarrow{PQ}\times \vec{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& -1\\ 1& 3& 2\end{array}\right|$

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

The equation of the plane passing through $P(0,4,3)$ with normal vector $\vec{n}=(1,-3,4)$ is:
$1(x-0)-3(y-4)+4(z-3)=0$
$x-3y+12+4z-12=0$
$x-3y+4z=0$

Comparing this with the given equation $ax+by+cz+d=0$ (where $a\ge 0$), we get:
$a=1$, $b=-3$, $c=4$, $d=0$

The value of $(a+b+c+d)$ is:
$1-3+4+0=2$

Answer: $2$