The line which contains all points `(x, y, z)` which are of the form `(x, y, z)=(2, -2, 5)+lambda(1, -3, 2)` intersects the plane `2x-3y+4z=163` at P and intersects the YZ-plane at Q. If the distance PQ is `asqrt(b)`, where `a,binN and agt3`, then `(a+b)` is equalto

Options

A. (a)`23`
B. (b)`95`
C. (c)`27`
D. (d)None of these

Answer

To solve the problem step-by-step, we will follow the outlined process to find the intersection points and the distance between them.

### Step 1: Define the line and the plane
The line is given in the parametric form:
\[ (x, y, z) = (2, -2, 5) + \lambda(1, -3, 2) \]
This can be expressed as:
\[ x = 2 + \lambda, \quad y = -2 - 3\lambda, \quad z = 5 + 2\lambda \]
The equation of the plane is:
\[ 2x - 3y + 4z = 163 \]

### Step 2: Substitute the line equations into the plane equation
We substitute the parametric equations of the line into the plane equation:
\[ 2(2 + \lambda) - 3(-2 - 3\lambda) + 4(5 + 2\lambda) = 163 \]
Expanding this:
\[ 4 + 2\lambda + 6 + 9\lambda + 20 + 8\lambda = 163 \]
Combining like terms:
\[ 4 + 6 + 20 + (2\lambda + 9\lambda + 8\lambda) = 163 \]
\[ 30 + 19\lambda = 163 \]
Now, solving for \(\lambda\):
\[ 19\lambda = 163 - 30 \]
\[ 19\lambda = 133 \quad \Rightarrow \quad \lambda = \frac{133}{19} = 7 \]

### Step 3: Find the coordinates of point P
Substituting \(\lambda = 7\) back into the parametric equations:
\[ x_P = 2 + 7 = 9 \]
\[ y_P = -2 - 3(7) = -2 - 21 = -23 \]
\[ z_P = 5 + 2(7) = 5 + 14 = 19 \]
Thus, the coordinates of point \(P\) are:
\[ P(9, -23, 19) \]

### Step 4: Find the intersection point Q with the YZ-plane
The YZ-plane is defined by \(x = 0\). Setting the equation for \(x\) from the line to zero:
\[ 2 + \lambda = 0 \quad \Rightarrow \quad \lambda = -2 \]
Now substituting \(\lambda = -2\) into the parametric equations:
\[ x_Q = 0 \]
\[ y_Q = -2 - 3(-2) = -2 + 6 = 4 \]
\[ z_Q = 5 + 2(-2) = 5 - 4 = 1 \]
Thus, the coordinates of point \(Q\) are:
\[ Q(0, 4, 1) \]

### Step 5: Calculate the distance PQ
Using 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 \(Q\):
\[ PQ = \sqrt{(0 - 9)^2 + (4 - (-23))^2 + (1 - 19)^2} \]
Calculating each term:
\[ = \sqrt{(-9)^2 + (4 + 23)^2 + (-18)^2} \]
\[ = \sqrt{81 + 27^2 + 324} \]
Calculating \(27^2\):
\[ = \sqrt{81 + 729 + 324} \]
\[ = \sqrt{1134} \]
This can be simplified:
\[ = \sqrt{9 \times 126} = 3\sqrt{126} \]

### Step 6: Identify \(a\) and \(b\)
We have \(PQ = a\sqrt{b}\) where \(a = 3\) and \(b = 126\). Since \(a\) must be greater than 3, we can take \(a = 9\) and \(b = 14\) (as \(126 = 9 \times 14\)).

### Final Step: Calculate \(a + b\)
Thus, \(a + b = 9 + 14 = 23\).

### Conclusion
The final answer is:
\[ \boxed{23} \]