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Step 1: Parametric Representation of Line

Let the given line be represented in parametric form. We introduce a parameter $\lambda$ such that:

$\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} = \lambda$

From this, we can express the coordinates of a general point P on the line as:

$x = \lambda - 5$

$y = 4\lambda - 3$

$z = -9\lambda + 6$

So, the coordinates of point P are $(\lambda - 5, 4\lambda - 3, -9\lambda + 6)$.

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Step 2: Distance Calculation

We are given that the distance between point P and point Q(2, 4, -1) is 7 units. We use the distance formula:

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substituting the coordinates of P and Q, we get:

$7 = \sqrt{(\lambda - 5 - 2)^2 + (4\lambda - 3 - 4)^2 + (-9\lambda + 6 - (-1))^2}$

$7 = \sqrt{(\lambda - 7)^2 + (4\lambda - 7)^2 + (-9\lambda + 7)^2}$

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Step 3: Squaring and Simplifying

Squaring both sides of the equation, we get:

$49 = (\lambda - 7)^2 + (4\lambda - 7)^2 + (-9\lambda + 7)^2$

$49 = (\lambda^2 - 14\lambda + 49) + (16\lambda^2 - 56\lambda + 49) + (81\lambda^2 - 126\lambda + 49)$

$49 = 98\lambda^2 - 196\lambda + 147$

$0 = 98\lambda^2 - 196\lambda + 98$

Dividing by 98, we get:

$0 = \lambda^2 - 2\lambda + 1$

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Step 4: Solving for λ

The quadratic equation is:

$\lambda^2 - 2\lambda + 1 = 0$

$(\lambda - 1)^2 = 0$

Therefore, $\lambda = 1$.

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Step 5: Finding Coordinates of P

Substitute $\lambda = 1$ into the coordinates of P:

$x = 1 - 5 = -4$

$y = 4(1) - 3 = 1$

$z = -9(1) + 6 = -3$

So, the coordinates of point P are $(-4, 1, -3)$.

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Step 6: Finding the Equation of Line PQ

The equation of the line joining two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Here, $P(-4, 1, -3)$ and $Q(2, 4, -1)$. Substituting these values, we get:

$\frac{x - (-4)}{2 - (-4)} = \frac{y - 1}{4 - 1} = \frac{z - (-3)}{-1 - (-3)}$

$\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}$

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\r\n Final Answer: P(-4, 1, -3) and Equation of line PQ: $\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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