Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The plane contains the line $\frac{x+1}{4}=\frac{y-2}{-1}=\frac{z+5}{3}$, so it passes through the point $(-1,2,-5)$ and is parallel to the vector ${\vec{v}}_1=4\hat{i}-\hat{j}+3\hat{k}$.

The plane is also parallel to the vector ${\vec{v}}_2=\hat{i}+\hat{j}-\hat{k}$.

The normal vector $\vec{n}$ to the plane is given by the cross product of ${\vec{v}}_1$ and ${\vec{v}}_2$:

$\vec{n}={\vec{v}}_1\times {\vec{v}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 3\\ 1& 1& -1\end{array}\right|$

$\vec{n}=\hat{i}(1-3)-\hat{j}(-4-3)+\hat{k}(4-(-1))=-2\hat{i}+7\hat{j}+5\hat{k}$

The equation of the plane passing through $(-1,2,-5)$ with normal vector $-2\hat{i}+7\hat{j}+5\hat{k}$ is:

$-2(x+1)+7(y-2)+5(z+5)=0$

$-2x-2+7y-14+5z+25=0$

$-2x+7y+5z+9=0$

$2x-7y-5z=9$

Comparing this with the given equation $ax+by+cz=9$, we get:

$a=2$, $b=-7$, $c=-5$

Therefore, the value of $a+b-c$ is:

$a+b-c=2+(-7)-(-5)=2-7+5=0$

Answer: $0$