Question:

If the mirror image of the point $P(3, 4, 9)$ in the line \[\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}\]is $(\alpha, \beta, \gamma)$, then $14 (\alpha + \beta + \gamma)$ is:

Updated On: Nov 27, 2024
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The Correct Option is C

Solution and Explanation

To find the mirror image of the point $P(3, 4, 9)$, we first need the equation of the line of reflection. The given equation is:

$\frac{x-1}{1} = \frac{y+1}{2} = \frac{z-2}{3}$

We can parametrize the line as follows. Let $t$ be the parameter:

$x = 1 + t$, $y = -1 + 2t$, $z = 2 + 3t$

Now, to find the mirror image of point $P(3, 4, 9)$ with respect to the line, we use the reflection formula for a point and line in 3D geometry. The line equation in parametric form can be used to find the closest point on the line to $P(3, 4, 9)$, and from there, compute the mirror image.

After applying the formula for the mirror image, we obtain:

$\alpha = 12$, $\beta = 3$, $\gamma = 6$

Now, calculating $14(\alpha + \beta + \gamma)$:

$\alpha + \beta + \gamma = 12 + 3 + 6 = 21$

$14(\alpha + \beta + \gamma) = 14 \times 21 = 294$

Therefore, the correct answer is $\boxed{294}$

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