Let the image of the point P(1,2,3) in the line L: x-63=y-12=z-23 be Q.

Question

Let the image of the point $P(1,2,3)$ in the line $L:\ \dfrac{x-6}{3}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$ be $Q$. Let $R(\alpha,\beta,\gamma)$ be a point that divides internally the line segment $PQ$ in the ratio $1:3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to:

Answer: 125

Accepted Answer

Since $R$ divides $PQ$ in ratio $1:3$, $R$ is the midpoint of $P$ and the foot of perpendicular $P'$ from $P$ onto $L$.

$P'(3\lambda+6,\ 2\lambda+1,\ 3\lambda+2)$; perpendicularity:
$$(3\lambda+5)(3)+(2\lambda-1)(2)+(3\lambda-1)(3)=0 \implies 22\lambda+5=0 \implies \lambda=-\frac{5}{11}.$$
$$P'=\left(\frac{51}{11},\frac{1}{11},\frac{7}{11}\right).$$
Midpoint of $P,P'$:
$$R=\left(\frac{62}{22},\frac{23}{22},\frac{40}{22}\right)=(\alpha,\beta,\gamma).$$
$$22(\alpha+\beta+\gamma)=62+23+40=125.$$

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