# 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 the area of parallelogram created by two adjacent vectors, a and b.

## 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 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