If the volume of a parallelopiped, whose coterminous edges are given by the vectors a⃗ =iˆ+jˆ+nkˆ ,  b⃗ =2iˆ+ 4jˆ− nkˆ and,c⃗ =iˆ+njˆ+3kˆ (n≥0) is 158 cubic units, then :

The volume of a parallelepiped formed by vectors a, b, and c is given by the scalar triple product of these vectors, which is the absolute value of the determinant formed by the components of the vectors.

The vectors given are:

a = i + j + nk b = 2i + 4j - nk c = i + nj + 3k

We can write these vectors in matrix form as follows:

| 1 1 n | | 2 4 -n | | 1 n 3 |

The scalar triple product (which gives the volume of the parallelepiped) is the determinant of this matrix.

So, we calculate the determinant as follows:

= 1*(43 - (-n)n) - 1(23 - (-n)1) + n(2n - 41) = 12 + n^2 - 6 + n + 2n^2 - 4 = 2n^2 + n + 8

Given that the volume of the parallelepiped is 158 cubic units, we can set up the equation:

2n^2 + n + 8 = 158 2n^2 + n - 150 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 2, b = 1, and c = -150. We can solve for n using the quadratic formula:

n = [-b ± sqrt(b^2 - 4ac)] / (2a) n = [-1 ± sqrt(1 - 42(-150))] / (2*2) n = [-1 ± sqrt(1 + 1200)] / 4 n = [-1 ± sqrt(1201)] / 4

Since n is non-negative (n ≥ 0), we take the positive root:

n = [-1 + sqrt(1201)] / 4 n = (sqrt(1201) - 1) / 4

So, the value of n that satisfies the given conditions is (sqrt(1201) - 1) / 4.