If the equation of a plane P, passing through the intersection of the planes, $$x + 4y - z + 7 = 0$$ and $$3x + y + 5z = 8$$ is $$ax + by + 6z = 15$$ for some $$a, b \in R$$, then the distance of the point (3, 2, -1) from the plane P is __________

We have two given planes

$$x+4y-z+7=0 \qquad\text{and}\qquad 3x+y+5z-8=0.$$

The standard result is: “Every plane that passes through the line of intersection of two planes $$P_1=0$$ and $$P_2=0$$ can be written as $$P_1+\lambda P_2=0$$, where $$\lambda\in\mathbb R.$$”

So the required plane can be written as

$$\bigl(x+4y-z+7\bigr)+\lambda\bigl(3x+y+5z-8\bigr)=0.$$

Expanding term-by-term, we obtain

$$\bigl(1+3\lambda\bigr)x+\bigl(4+\lambda\bigr)y+\bigl(-1+5\lambda\bigr)z+\bigl(7-8\lambda\bigr)=0.$$

Now it is given that the same plane can also be expressed as

$$ax+by+6z=15.$$

To compare both forms, it is convenient to rewrite the second one with all terms on the left:

$$ax+by+6z-15=0.$$

If two linear equations represent the same plane, all corresponding coefficients must be proportional. Therefore, there exists some non-zero constant $$k$$ such that

$$k\bigl(1+3\lambda\bigr)=a,$$ $$k\bigl(4+\lambda\bigr)=b,$$ $$k\bigl(-1+5\lambda\bigr)=6,$$ $$k\bigl(7-8\lambda\bigr)=-15.$$

We use the last two equalities to determine $$\lambda$$ and $$k$$. First, form their ratio to eliminate $$k$$:

$$\frac{k(-1+5\lambda)}{k(7-8\lambda)}=\frac{6}{-15}=-\frac{2}{5}.$$

This gives the linear equation

$$\frac{-1+5\lambda}{7-8\lambda}=-\frac{2}{5}.$$

Cross-multiplying yields

$$5(-1+5\lambda)=-2(7-8\lambda).$$

Simplifying both sides,

$$-5+25\lambda = -14+16\lambda.$$

Bringing like terms together,

$$25\lambda-16\lambda = -14+5 \quad\Longrightarrow\quad 9\lambda = -9.$$

Hence

$$\lambda = -1.$$

Now substitute $$\lambda=-1$$ into $$k(7-8\lambda)=-15$$ to find $$k$$:

$$7-8(-1)=7+8=15,$$

so

$$k\cdot15=-15 \quad\Longrightarrow\quad k=-1.$$

Finally we compute the coefficients $$a$$ and $$b$$:

$$a = k(1+3\lambda)=(-1)\bigl(1+3(-1)\bigr)=(-1)(-2)=2,$$

$$b = k(4+\lambda)=(-1)\bigl(4+(-1)\bigr)=(-1)(3)=-3.$$

Thus the explicit equation of the plane is

$$2x-3y+6z=15 \quad\Longleftrightarrow\quad 2x-3y+6z-15=0.$$

Now we calculate the perpendicular distance of the point $$P(3,2,-1)$$ from this plane. The distance formula for a point $$(x_0,y_0,z_0)$$ from the plane $$A x+B y+C z+D=0$$ is

$$\text{Distance}=\dfrac{\lvert A x_0+B y_0+C z_0+D\rvert}{\sqrt{A^{2}+B^{2}+C^{2}}}.$$

Here $$A=2,\;B=-3,\;C=6,\;D=-15$$ and $$(x_0,y_0,z_0)=(3,2,-1).$$ Substituting these values,

$$\begin{aligned} \lvert A x_0+B y_0+C z_0+D\rvert &=\bigl|\,2\cdot3+(-3)\cdot2+6\cdot(-1)-15\,\bigr|\\[4pt] &=\bigl|\,6-6-6-15\,\bigr|\\[4pt] &=\lvert -21\rvert=21. \end{aligned}$$

Also

$$\sqrt{A^{2}+B^{2}+C^{2}} =\sqrt{2^{2}+(-3)^{2}+6^{2}} =\sqrt{4+9+36} =\sqrt{49}=7.$$

Therefore, the required distance is

$$\frac{21}{7}=3.$$

So, the answer is $$3$$.