Answer:

It has one solution
Step-by-step explanation:
![14x + 12] = 0\\\\14x+12=0\\\mathrm{Subtract\:}12\mathrm{\:from\:both\:sides}\\14x+12-12=0-12\\\\Simplify\\14x=-12\\\\\mathrm{Divide\:both\:sides\:by\:}14\\\frac{14x}{14}=\frac{-12}{14}\\\\Simplify\\x=-\frac{6}{7}](https://tex.z-dn.net/?f=14x%20%2B%2012%5D%20%3D%200%5C%5C%5C%5C14x%2B12%3D0%5C%5C%5Cmathrm%7BSubtract%5C%3A%7D12%5Cmathrm%7B%5C%3Afrom%5C%3Aboth%5C%3Asides%7D%5C%5C14x%2B12-12%3D0-12%5C%5C%5C%5CSimplify%5C%5C14x%3D-12%5C%5C%5C%5C%5Cmathrm%7BDivide%5C%3Aboth%5C%3Asides%5C%3Aby%5C%3A%7D14%5C%5C%5Cfrac%7B14x%7D%7B14%7D%3D%5Cfrac%7B-12%7D%7B14%7D%5C%5C%5C%5CSimplify%5C%5Cx%3D-%5Cfrac%7B6%7D%7B7%7D)
5/12 is the simplest form that it can be simplified to.
Is there any option to choose from ?
If so show them and I’ll probably be able to answer .
A) There are a number of ways to compute the determinant of a 3x3 matrix. Since k is on the bottom row, it is convenient to compute the cofactors of the numbers on the bottom row. Then the determinant is ...
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.