Question

The equation of the planes parallel to the plane x - 2y + 2z - 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b - d) = K(c - a), then the positive value of K is __________.

Hint:

Distance from $(x_1, y_1, z_1)$ to plane $ax+by+cz+d=0$ is $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$.

Answer 1

Correct Answer: 4

Step 1: Parallel plane: $x - 2y + 2z + d = 0$.
Step 2: Distance from $(1, 2, 3)$ is 1: $\frac{|1 - 2(2) + 2(3) + d|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$.
Step 3: $\frac{|1 - 4 + 6 + d|}{3} = 1 \implies |3+d| = 3$.
Step 4: $3+d = 3 \implies d=0$ or $3+d = -3 \implies d=-6$.
Step 5: Let's take $x - 2y + 2z - 6 = 0$. Here $a=1, b=-2, c=2, d=-6$.
Step 6: $b-d = -2 - (-6) = 4$.
Step 7: $c-a = 2 - 1 = 1$.
Step 8: $4 = K(1) \implies K = 4$.