# Find the coordinates of the foot of the perpendicular drawn from the point (2, 3, 7) to the plane 3x – y – z = 7. Also, find the length of the perpendicular.

Let point P = (2, 3, 7) and Q be the foot of the perpendicular drawn from P to the plane 3x – y – z = 7.

Direction ratios of PQ are proportional to 3, –1, –1 as PQ is normal to the plane.

Recall the equation of the line passing through (x1, y1, z1) and having direction ratios proportional to l, m, n is given by Here, (x1, y1, z1) = (2, 3, 7) and (l, m, n) = (3, –1, –1)

Hence, the equation of PQ is  x = 3α + 2, y = 3 – α, z = 7 – α

Let Q = (3α + 2, 3 – α, 7 – α).

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

3(3α + 2) – (3 – α) – (7 – α) = 7

9α + 6 – 3 + α – 7 + α = 7

11α – 4 = 7

11α = 11

α = 1

We have Q = (3α + 2, 3 – α, 7 – α)

Q = (3×1 + 2, 3 – 1, 7 – 1)

Q = (5, 2, 6)

Using the distance formula, we have    Thus, the required foot of perpendicular is (5, 2, 6) and the length of the perpendicular is units.

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Find the coordinate of the point P where the line through and crosses the plane passing through three points and Also, find the ratio in which P divides the line segment AB.

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