\( 2 x+2 y+z-7=0,2 x-y+3 z-3=0,4 x+5 y-2 z-12=0 \) tekisliklarning kesishish nuqtasi va \( M(0 ; 3 ; 0), N(1: 1 ; 1) \) nuqtalardan o'tuvchi tekislik tenglamasi topilsin.

Uchta tekislikning tenglamalari quyidagicha berilgan: \[ \begin{cases} 2x + 2y + z = 7 \quad \text{(1)} \\ 2x - y + 3z = 3 \quad \text{(2)} \\ 4x + 5y - 2z = 12 \quad \text{(3)} \end{cases} \] **Kesishish nuqtasini topish:** 1. **(1) va (2) tenglamalarni yechish:** \[ \begin{align*} (2x - y + 3z) - (2x + 2y + z) &= 3 - 7 \\ -3y + 2z &= -4 \quad \text{(4)} \end{align*} \] 2. **(3) tenglamani (1) tenglamaning ikki baravari bilan kamaytirish:** \[ \begin{align*} (4x + 5y - 2z) - 2(2x + 2y + z) &= 12 - 14 \\ y - 4z &= -2 \quad \text{(5)} \end{align*} \] 3. **(5) dan y ni ifodalash va (4) ga qo'yish:** \[ y = 4z - 2 \] \[ -3(4z - 2) + 2z = -4 \\ -12z + 6 + 2z = -4 \\ -10z = -10 \\ z = 1 \] \[ y = 4(1) - 2 = 2 \] 4. **(1) tenglamadan x ni topish:** \[ 2x + 2(2) + 1 = 7 \\ 2x + 5 = 7 \\ 2x = 2 \\ x = 1 \] Demak, uchta tekislikning kesishish nuqtasi \( (1, 2, 1) \) hisoblanadi. **Tekislik tenglamasini topish:** Berilgan nuqtalar: - \( M(0, 3, 0) \) - \( N(1, 1, 1) \) - Kesishish nuqtasi \( (1, 2, 1) \) 1. **Vektorlarni aniqlash:** \[ \vec{A} = M - (1, 2, 1) = (-1, 1, -1) \] \[ \vec{B} = N - (1, 2, 1) = (0, -1, 0) \] 2. **Normal vektorni topish (kross mahsuloti):** \[ \vec{n} = \vec{A} \times \vec{B} = (-1, 1, -1) \times (0, -1, 0) = (-1)\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} = (-1, 0, 1) \] 3. **Tekislik tenglamasi:** \[ -1(x - 1) + 0(y - 2) + 1(z - 1) = 0 \\ -x + z = 0 \\ x - z = 0 \] Demak, \( M \) va \( N \) nuqtalardan o'tuvchi tekislik tenglamasi: \[ x - z = 0 \]