Question

The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane 3x + 4y + 5z = 10 is:

• A. (-2, 6, -4)
• B. (2, -6, -4)
• C. (2, 6, -4)
• D. (2, 6, 4)

Answer 1

Answer: (C)

Let $\frac{x+1}{1}=\frac{y+3}{3}=\frac{-z+2}{2}=k$
Thus any point on this line will have coordinates,
x = k - 1, y = 3k - 3, z = - 2k + 2
This line intersects the plane
3x + 4y + 5z = 10. Hence value of x, y, z must satisfy equation of plane.
$\Rightarrow$ 3(k - 1) + 4(3k - 3) + 5 (-2k + 2) = 10
$\Rightarrow$ 3k - 3 + 12k - 12 + 10 - 10k = 10
$\Rightarrow$ 5k - 5 = 10
$\Rightarrow$ k = 3
$\therefore$ The point of intersection is (x, y, z)
$\Rightarrow$ x = 3 - 1, y = 3 $\times$ 3 - 3, z = - 2 $\times$ 3 + 2
$\Rightarrow$ (x, y, z) = (2, 6 - 4)