Given a point $(1, -5, 6)$ is reflected about the plane having equation $-2x+7y+9z=4$. What will the point of reflection about the plane?
a.) $\left(\frac{98}{67}, \frac{-389}{67}, \frac{-185}{67} \right)$
b.) $\left(\frac{93}{67}, \frac{-426}{67}, \frac{285}{67} \right)$
c.) $\left(\frac{-80}{67}, \frac{426}{67}, \frac{288}{67} \right)$
d.) None
I know that the point of reflection will at a distance from the plane equal to distance of the point $(1, -5, 6)$ from the plane but can't proceed further.
I studied math up to 12th. Thanks!
suppose the reflection of $A=(1, -5, 6)$ on the plane $$-2x+7y+9z = 4\tag 1$$ is $$A' = (1 - 4t, -5+14t, 6+18t)\tag 2.$$ then the midpoint $$(A+A')/2 = (1 - 2t, -5+7t, 6+9t) $$ must be on the plane $(1).$ therefore $t$ must satisfy $$-2(1-2t)+7(-5+7t)+9(6+9t) = 4\to 134 t = -7, t=-7/134.$$ now putting this value of $t$ in $(2)$ will give us the required point.