Any plane parallel to 2x-3y+5z+7=0 is given by 2x-3y+5z+d=0

Since it passes through (3, 4, -1)

2x3-3x4+5x(-1)+d=0

d=11

required equation is

2x-3y+5z+11=0

Given: The given line is 3x + 5y = 4. 
then the slope of the given line = -3/5

Formula to be used: The product of slopes of two perpendicular lines = - 1.

The slope of this line is – 3/5

∴ the slope of the perpendicular line = 5/3 
now we have a slope of the line and the point (2,3) from where the line is passing 
then the equation of line will be 

The equation of the line can be written in the form

(Y-y1)=m(X-x1)
(y-3)=(5/3)(x-2)
3y-9=5x-10
5x-3y=1

The required equation of the line is

5x-3y=1

Plane parallel to given plane 2x-y+3z+d=0

Since it passes through (1,4,-2)

d=8

required equation is 2x-y+3z+8=0

Given: Point (3,-5) and line 3x – 4y = 27

To find: The distance of the point (3, -5) from the line 3x – 4y = 27

Formula used:

We know that the distance between a point P(m,n) and a line ax + by + c = 0 is given

by,

The distance of the point (3,-5) from the line 3x – 4y = 27 is 2/5 unit.

The given equations are x + y = 6 …(i)

x – 3y = 2 …(ii) and

5x – 3y + 2 = 0 or 5x – 3y = -2 …(iii)

Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC
Firstly, we solve the equation (i) and (ii)

x + y = 6 …(i)

x – 3y = 2 …(ii)

Subtracting eq. (ii) from (i), we get

x + y – x + 3y = 6 – 2

⇒ 4y = 4

⇒ y = 1

Putting the value of y = 1 in eq. (i), we get

x + 1 = 6

⇒ x = 5

Thus, AB and BC intersect at (5, 1)

Now, we solve eq. (ii) and (iii)

x – 3y = 2 …(ii)

5x – 3y = -2 …(iii)

Subtracting eq. (ii) from (iii), we get

5x – 3y – x + 3y = – 2 – 2

⇒ 4x = - 4

⇒ x = -1

Putting the value of x = -1 in eq. (ii), we get

– 1 – 3y = 2

⇒ -3y = 2 + 1

⇒ -3y = 3

⇒ y = -1

Thus, BC and AC intersect at (-1, -1)

Now, we solve eq. (iii) and (i)

5x – 3y = -2 …(iii)

x + y = 6 …(i)

From eq. (i), we get

x = 6 – y

Putting the value of x in eq. (iii), we get

5(6 – y) – 3y = -2

⇒ 30 – 5y – 3y = -2

⇒ 30 – 8y = -2

⇒ -8y = -32

⇒ y = 4

Putting the value of y = 4 in eq. (i), we get

x + 4 = 6

⇒ x = 6 – 4

⇒ x = 2

Thus, AC and AB intersect at (2, 4) So, vertices of triangle ABC are: (5, 1), (-1, -1) and (2, 4)

The given points are A(-3,2), B(-5,-5), C(2,-3), D(4,4)

Thus, diagonal AC is not equal to diagonal BD.

Therefore ABCD is a quadrilateral with equal sides and unequal diagonals.

Hence, ABCD a rhombus.

Area of a rhombus=1/2× (product of diagonals)

Two vertices of ∆ABC are A(1, -6) and B(-5, 2)

Let the third vertex be C(a, b).

Then the coordinates of its centroid are

a= -2 and b=7

Therefore, the third vertex of ∆ABC is C(-2,7).

P(1, 4), Q(3,y) and R(-3, 16) are the given points. Then

(x₁=1, y₁=4), (x₂=3, y₂=y), (x₃=-3, y₃=16)

It is given that points P,Q and R are collinear.

Therefore,

x₁(y₂-y₃)+ x₂(y₃-y₁)+ x₃(y₁-y₂)=0

⇒ 1(y-16)+3(16-4)-3(4-y)=0

⇒ 1(y-16)+3(12)-3(4-y)=0

⇒ y-16+36-12+3y=0

⇒ 8+4y=0

⇒ y=-8/4=-2

so, for y=-2 points are collinear.