If P(5, 4, a) and Q(-1, 2, -2) are two points | vedclass

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(A) Given the points $P(5, 4, a)$ and $Q(-1, 2, -2)$ with distance $PQ = 7$.Using the distance formula: $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 -

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$-5, 1$

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(A) Given the points $P(5, 4, a)$ and $Q(-1, 2, -2)$ with distance $PQ = 7$.Using the distance formula: $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.Substituting the values: $7^2 = (-1 - 5)^2 + (2 - 4)^2 + (-2 - a)^2$.$49 = (-6)^2 + (-2)^2 + (-(2 + a))^2$.$49 = 36 + 4 + (a + 2)^2$.$49 = 40 + (a + 2)^2$.$(a + 2)^2 = 49 - 40 = 9$.Taking the square root on both sides: $a + 2 = \pm 3$.Case $1$: $a + 2 = 3 \implies a = 1$.Case $2$: $a + 2 = -3 \implies a = -5$.Therefore,the values of $a$ are $-5$ and $1$.

If $P(5, 4, a)$ and $Q(-1, 2, -2)$ are two points such that $PQ = 7$,then the values of $a$ are ..........

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