Statement-I A point on the straight line `2x+3y-4z=5 and 3x-2y+4z=7` can be determined by taking x=k and then solving the two for equation for y and z, where k is any real number.
Statement-II If `c'nekc`, then the straight line `ax+by+cz+d=0, Kax+Kby+c'z+d'=o` does not intersect the plane `z=alpha`, where `alpha` is any real number.

Options

A. Statement I is true, Statement II is also true, Statement-II is the correct explanation of Statement-I.
B. Statement-I is true, Statement-II is also true, Statement-II is not the correct explanation of Statement-I.
C. Statement-I is true, Statement-II is false.
D. Statement-I is false, Statement -II is true.

Answer

To solve the problem, we need to analyze both statements provided.

### Step-by-Step Solution:

**Statement I:**
We are given two equations of planes:
1. \(2x + 3y - 4z = 5\) (Equation 1)
2. \(3x - 2y + 4z = 7\) (Equation 2)

To find a point on the line of intersection of these two planes, we can set \(x = k\) (where \(k\) is any real number) and solve for \(y\) and \(z\).

1. Substitute \(x = k\) into Equation 1:
\[ 2k + 3y - 4z = 5 \quad \Rightarrow \quad 3y - 4z = 5 - 2k \quad \text{(Equation 3)} \]

2. Substitute \(x = k\) into Equation 2:
\[ 3k - 2y + 4z = 7 \quad \Rightarrow \quad -2y + 4z = 7 - 3k \quad \text{(Equation 4)} \]

Now we have two equations (Equation 3 and Equation 4) with two variables \(y\) and \(z\).

3. We can solve these two equations simultaneously to find \(y\) and \(z\) in terms of \(k\).

From Equation 3:
\[ 3y = 4z + 5 - 2k \quad \Rightarrow \quad y = \frac{4z + 5 - 2k}{3} \]

Substituting \(y\) from this equation into Equation 4:
\[ -2\left(\frac{4z + 5 - 2k}{3}\right) + 4z = 7 - 3k \]

4. Multiply through by 3 to eliminate the fraction:
\[ -2(4z + 5 - 2k) + 12z = 21 - 9k \]
\[ -8z - 10 + 4k + 12z = 21 - 9k \]
\[ 4z + 13k = 31 \quad \Rightarrow \quad z = \frac{31 - 13k}{4} \]

5. Substitute \(z\) back into the equation for \(y\):
\[ y = \frac{4\left(\frac{31 - 13k}{4}\right) + 5 - 2k}{3} \]
\[ y = \frac{31 - 13k + 5 - 2k}{3} = \frac{36 - 15k}{3} = 12 - 5k \]

Thus, we have:
- \(x = k\)
- \(y = 12 - 5k\)
- \(z = \frac{31 - 13k}{4}\)

This shows that a point on the line can indeed be determined by taking \(x = k\) and solving for \(y\) and \(z\).

**Conclusion for Statement I: True.**

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**Statement II:**
We are given a condition that if \(c' \neq kc\), then the lines represented by:
1. \(ax + by + cz + d = 0\)
2. \(kax + kby + c'z + d' = 0\)

do not intersect the plane \(z = \alpha\) (where \(\alpha\) is any real number).

To analyze this, we can express the equations in terms of \(z\):
1. From the first equation:
\[ z = -\frac{ax + by + d}{c} \]
2. From the second equation:
\[ z = -\frac{kax + kby + d'}{c'} \]

For these two lines to not intersect the plane \(z = \alpha\), it must be that the two expressions for \(z\) yield different values for all \(x\) and \(y\).

Given the condition \(c' \neq kc\), we can conclude that the slopes of the lines in the \(z\) direction are different, which means they will not intersect the plane \(z = \alpha\).

**Conclusion for Statement II: True.**

### Final Conclusion:
Both statements are true, and Statement II is a correct explanation of Statement I.