Let the equation of plane passing through the line of intersection of the planes x+2 y+a z…

Question

Let the equation of plane passing through the line of intersection of the planes $x+2 y+a z=2$ and $x-y+z=3$ be $5 x-11 y+b z=6 a-1$. For $$c \in \mathbb{Z}$$, if the distance of this plane from the point $(a,-c, c)$

Accepted Answer

Given the equation of the plane passing through the intersection of the two given planes:

$$P: (x + 2y + az - 2) + \lambda(x - y + z - 3) = 0$$

$$\Rightarrow x(\lambda+1)+y(2-\lambda)+z(a+\lambda)-2-3 \lambda=0$$

This is the same as the given equation $5x - 11y + bz = 6a - 1$.

Now, comparing the coefficients of the corresponding variables in both equations:

$$\frac{\lambda+1}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$$

Solving for $$\lambda$$:

$$-11\lambda -11 = 10 - 5\lambda$$

$$6\lambda = -21 \Rightarrow \lambda = -\frac{7}{2}$$

Now, substituting the value of $$\lambda$$ back into the equations:

$$\frac{2-\lambda}{-11} = \frac{2+3\lambda}{6a-1} \Rightarrow \frac{2+\frac{7}{2}}{-11} = \frac{2-\frac{21}{2}}{6a-1}$$

From this equation, we find the value of a :

$$6a - 1 = 17 \Rightarrow a = 3$$

Now, substituting the value of $a$ and $$\lambda$$ into the equation:

$$\frac{2-\lambda}{-11} = \frac{a+\lambda}{b} \Rightarrow -\frac{1}{2} = \frac{3 - \frac{7}{2}}{b}$$

$$\Rightarrow -\frac{b}{2} = -\frac{1}{2} \Rightarrow b = 1$$

Therefore, the point $$(a, -c, c) \equiv (3, -c, c)$$.

The given distance is $$\frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}}$$.

The plane is: $5x - 11y + z = 17$.

Now, let's find the distance:

$$\left|\frac{15 + 11c + c - 17}{\sqrt{147}}\right| = \frac{2}{\sqrt{3}}$$

$$\Rightarrow c = -1, \frac{4}{3}$$

Since $$c \in \mathbb{Z}$$, we have $c = -1$.

Therefore, $$\frac{a+b}{c} = \frac{3+1}{-1} = -4$$.

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