Let the image of the point P(1,2,6) in the plane passing through the points A(1,2,0),B(1,4…

Question

Let the image of the point $P(1,2,6)$ in the plane passing through the points $A(1,2,0),B(1,4,1)$ and $C(0,5,1)$ be $Q(\alpha ,\beta ,\gamma )$. Then $({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2})$ is equal to :

Accepted Answer

Equation of plane passing through the points $A(1,2$, $0), B(1,4,1)$ and $C(0,5,1)$ is

$$
\begin{aligned}
& \left|\[\begin{array}{ccc} x-1 & y-2 & z-0 \ 0 & 2 & 1 \ -1 & 3 & 1 \end{array}\end{aligned}$$\right|=0 \\
\& \Rightarrow x+y-2 z=3

\]

Now $Q(\alpha, \beta, \gamma)$ is the image of the point $P(1,2,6)$ in the plane $x+y-2 z-3=0$

$$
\[\begin{gathered} \therefore  \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2[1+2-2(6)-3]}{1^2+1^2+(-2)^2} \ \Rightarrow  \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4 \ \Rightarrow  \alpha=5, \beta=6, \gamma=-2 \end{gathered}$$
\]

$$
\text { Hence, } \alpha^2+\beta^2+\gamma^2=5^2+6^2+(-2)^2=65
$$

Solutions · 1