# Find the distance

Let P be the point with position vector and Q be the point of intersection of the given line and the plane.

We have the line equation as

Let the position vector of Q be. As Q is a point on this line, for some scalar α, we have

This point Q also lies on the given plane, which means this point satisfies the plane equation.

(2 + 3α)(1) + (–1 + 4α)(–1) + (2 + 12α)(1) = 5

2 + 3α + 1 – 4α + 2 + 12α = 5

11α + 5 = 5

11α = 0

α = 0

We have

Using the distance formula, we have

Thus, the required distance is 13 units.

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