Given:

2 oranges + 3 bananas + 4 apples = Rs.15 ----(1)
3 oranges + 2 bananas + 1 apple = Rs.10 ----(2)

To find:
Cost of 3 oranges + 3 bananas + 3 apples

Solution:
Let's first find the cost of 1 orange, 1 banana, and 1 apple.

From equation (1), we can write:
2 oranges + 3 bananas + 4 apples = Rs.15

Divide both sides by 2 to get the cost of 1 orange:
1 orange + 3/2 bananas + 2 apples = Rs.7.5

Subtracting the cost of 3 bananas and 2 apples from both sides, we get:
1 orange = Rs.7.5 - 3/2(Rs.10) = Rs.7.5 - Rs.15 = -Rs.7.5

This means that 1 orange has a negative cost, which is not possible.
So, we made a mistake in assuming that equation (1) gives us the cost of the fruits.

Let's try to find the cost of each fruit from equation (2):

3 oranges + 2 bananas + 1 apple = Rs.10

Divide both sides by 3 to get the cost of 1 orange:
1 orange + 2/3 bananas + 1/3 apple = Rs.10/3

Subtracting the cost of 2 bananas and 1/3 apple from both sides, we get:
1 orange = Rs.10/3 - 2/3(Rs.15) = Rs.10/3 - Rs.10 = -Rs.20/3

Again, we have a negative cost for 1 orange, which is not possible.

Let's try to find the cost of each fruit in a different way:

Let the cost of 1 orange be x, the cost of 1 banana be y, and the cost of 1 apple be z.
From equation (1), we get:
2x + 3y + 4z = 15

From equation (2), we get:
3x + 2y + z = 10

Multiplying equation (2) by 2 and subtracting it from equation (1) multiplied by 3, we get:
3(2x + 3y + 4z) - 2(3x + 2y + z) = 15 - 2(10)
6x + 9y + 12z - 6x - 4y - 2z = -5
5y + 10z = -5
y + 2z = -1 ----(3)

Now we can use equations (2) and (3) to solve for x, y, and z:

From equation (3), we get:
y = -1 - 2z

Substituting this in equation (2), we get:
3x + 2(-1 - 2z) + z = 10
3x - 4z = 12 ----(4)

Substituting y = -1 - 2z and y + 2z = -1 in equation (1), we get:
2x + 3(-1 - 2z) + 4