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# How do you find the Area of the Parallelogram with Vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3)?

We will be using the concept of cross-product of two adjacent vectors, a and b to find the area of the parallelogram when vertices given.

## Answer: The Area of the Parallelogram with Vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3) is √265.

Let's solve this step by step.

**Explanation:**

Given that, vertices k(1, 2, 3), l(1, 3, 6), m(3, 8, 6), and n(3, 7, 3)

Area of Parallelogram = |KL × KN|

KL = (1, 2, 3) - (1, 3, 6) = (0, 1, 3)

KN = (1, 2, 3) - (3, 7, 3) = (2, 5, 0)

Area of the parallelogram = cross product of two vectors, represented by the adjacent sides.

Area of Parallelogram = |(0, 1, 3) × (2, 5, 0)|

= |i(1 × 0 - 5 × 3) - j(0 × 0 - 2 × 3) + k(0 × 5 - 2 × 1)|

= |i(-15) - j(-6) + k(-2)|

= |-15i + 6j -2 k|

= √(-15)^{2} + (6)^{2} + (-2)^{2}

= √(225 + 36 + 4)

= √265

### Thus, the area of the parallelogram with vertices k(1,2,3), l(1,3,6), m(3,8,6), and n(3,7,3) is √265.

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