The distance between the point $(-1, -5, -10)$ and the point of intersection of the line $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-2}{-2}$ with the plane $x + y + z = 5$ is $13t$, then $t$ equals to
Step-by-Step Solution
Key Concept: Find the intersection point of a line with a plane by substituting the parametric form of the line into the plane equation.
Any point on the line is $(3t+2, 4t-1, 12t+2)$. If it lies on the given plane, then $(3t+2)-(4t-1)+(12t+2)=5$, which simplifies to $11t=0$ or $t=0$. The coordinates of the point of intersection of the given line and given plane are $(2,-1,2)$ and thus the required distance is $\sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}=\sqrt{13}$. Thus $13=13t$, so $t=1$.
Correct Answer: 1