For x,

2x+6 = 0

⇒ 2x = -6

⇒ X = -6/2

⇒ X = -3

For y,

4y -14 = 6y-4

⇒ 4y-6y = -4 +14

⇒ -2y = 10

⇒ Y = 10/(-2)

⇒ Y = -5

For z,

2z+8 = 12

⇒ 2z = 12-8

⇒ 2z = 4

⇒ Z = 2

Using elementary operations on a matrix;

⇒ Expanding along R2, gives both determinant values as 0.

= 10(3*13-0) – 17(0) + 1(0-3)

= 10(39) – 0 -3

=390-3

=387

Determinant of A

= 3(12*3 -0*4) – 8((-3*3) – 12*0) + 4((-3*4) – 12*12)

= 3(12) -8(-9) + 4( -12-144)

= 3*36 + 72 + 4*(-156)

= 108 + 72 – 624

= -444

For a square matrix A, the determinant of matrix is, when all the elements are multiplied by a constant k, becomes, kⁿ|A|

M₂₂ = (1*9)-(7*3) = 9-21 = -12

Cofactor of A22 = (4*(-14)) – (5*2) = -56 - 10 = -66

AB=BC

Therefore, Δ ABC is an isosceles triangle .....(3)

Also, AB² = 106 units .....(4)

BC² = 106 units .....(5)

AC² = 212 units .....(6)

From equations 4, 5 and 6, we have

AB² + BC² = AC²

So, it satisfies the Pythagoras theorem.

Δ ABC is right angled triangle .....(7)

From 3 and 7, we have

Δ ABC is an isosceles right angled triangle.

Let the equation of the required ellipse be

Coordinates of foci = (±3, 0) ...(ii)

We know that,

Coordinates of foci = (±c, 0) ...(iii)

∴ From eq. (ii) and (iii), we get

c = 3

We know that,

C² = a² – b²

⇒ (3)²= a² – b²

⇒ 9 = a²– b²

⇒ b²= a² – 9 ...(iv)

Given that ellipse passing through the points (4, 1)

So, point (4, 1) will satisfy the eq. (i)

Taking point (4, 1) where x = 4 and y = 1

Putting the values in eq. (i), we get

16a²– 144 + a² = a²(a² – 9)

⇒ 17a² – 144 = a⁴ – 9a²

⇒ a⁴– 9a² – 17a² + 144 = 0

⇒ a⁴ – 26a² + 144 = 0

⇒ a⁴ – 8a² – 18a² + 144 = 0

⇒ a²(a² – 8) – 18(a² – 8) = 0

⇒ (a² – 8)(a² – 18) = 0

⇒ a² – 8 = 0 or a² – 18 = 0

⇒ a² = 8 or a² = 18

If a² = 8 then

b²= 8– 9=- 1

Since the square of a real number cannot be negative. So, this is not possible

If a² = 18 then

b² = 18 – 9= 9

So, equation of ellipse if a2 = 18 and b2 = 9

Equation of the planes 2x+3y+4z=1 and 4x+6y+8z=12 or 2x+3y+4z = 6

We know that distance between ax+by+cz+d1=0 and ax+by+cz+d1=0 is