Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane is given by $\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}$, where $\vec{a}=\hat{k}$, $\vec{b}=-\hat{i}+\hat{j}$, and $\vec{c}=-\hat{i}+\hat{k}$.

The normal vector to the plane is $\vec{n}=\vec{b}\times \vec{c}=(-\hat{i}+\hat{j})\times (-\hat{i}+\hat{k})=\hat{i}+\hat{j}+\hat{k}$.

The equation of the plane is $\vec{r}\cdot \vec{n}=\vec{a}\cdot \vec{n}$, which gives $(x\hat{i}+y\hat{j}+z\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=\hat{k}\cdot (\hat{i}+\hat{j}+\hat{k})$.

$x+y+z=1$

Using the formula for the mirror image $(\alpha ,\beta ,\gamma )$ of a point $(x_1,y_1,z_1)$ in the plane $ax+by+cz+d=0$:

$\frac{\alpha -x_1}{a}=\frac{\beta -y_1}{b}=\frac{\gamma -z_1}{c}=-2\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$

Substituting the given values:

$\frac{\alpha -10}{1}=\frac{\beta -15}{1}=\frac{\gamma -20}{1}=-2\frac{10+15+20-1}{1^2+1^2+1^2}$

$\frac{\alpha -10}{1}=\frac{\beta -15}{1}=\frac{\gamma -20}{1}=-2\left(\frac{44}{3}\right)=-\frac{88}{3}$

From this, we get:

$\alpha =10-\frac{88}{3}$

$\beta =15-\frac{88}{3}$

$\gamma =20-\frac{88}{3}$

Adding these equations:

$\alpha +\beta +\gamma =45-3\left(\frac{88}{3}\right)=45-88=-43$

Therefore, $3(\alpha +\beta +\gamma )=3(-43)=-129$.

Answer: $-129$